Arithmetic of Hermitian Forms

Goro Shimura

Received: May 27, 2008

Communicated by Don Blasius

Abstract. We investigate the following two problems on a hermitian form Φ over an algebraic number field: (1) classification of Φ over the ring of algebraic integers; (2) hermitian Diophantine equations. The same types of problems for quadratic forms were treated in the author’s previous articles. Here we discuss the hermitian case. Problem (2) concerns an equation ξΦ ·^{t}ξ^{ρ} = Ψ, where Φ and Ψ represent hermitian forms. We connect the number of such ξ modulo a group of units with the class number and mass of the unitary group of a form Θ such that Φ ≈ Ψ ⊕ Θ.

2000 Mathematics Subject Classification: 11E39 (primary), 11E41, 11D09 (secondary)

Introduction

To explain Problems (1) and (2) of the abstract, we take a quadratic extension K of an algebraic number field F, a vector space V over K of dimension n, and a nondegenerate hermitian form φ : V × V → K with respect to the nontrivial automorphism ρ of K over F. We denote by d_{0}(φ) the coset of F^{×}/N_{K/F} (K^{×}) represented by (−1)^{n(n−1)/2} det(φ). It is classically known that n, d_{0}(φ), and the indices of φ at certain archimedean primes of F, satisfying a natural consistency condition, determine the isomorphism class of (V, φ), and vice versa. This classification does not answer, however, the question of classification over the ring of integers. To be precise, let 𝔯 denote the ring of algebraic integers in K and 𝔤 = F ∩ 𝔯; let 𝔡 be the different of K relative to F. We put

(0.1) ℌ_{n} = {Φ ∈ GL_{n}(K) | Φ = ^{t}Φ^{ρ}}.

We call a matrix Φ = (φ_{ij}) ∈ ℌ_{n} semi-integral if φ_{ij} ∈ 𝔡^{−1} and φ_{ii} ∈ 𝔤 for every i and j, which means that ∑_{i,j} φ_{ij}x_{i}x_{j}^{ρ} ∈ 𝔤 for every (x_{i})_{i=1}^{n} ∈ 𝔯^{n}. Further we call a semi-integral Φ reduced if the following condition is satisfied:

(R) If Φ = P Ψ ·^{t}P^{ρ} with a semi-integral Ψ and P = (p_{ij}) ∈ GL_{n}(K), p_{ij} ∈ 𝔯, then det(P) ∈ 𝔯^{×}.

These definitions are natural, but cover only a special class of (V, φ), as an 𝔯-lattice in V may not be isomorphic to 𝔯^{n}. In order to classify all 𝔤-valued hermitian forms, we have to define the genus of a form relative to an isomorphism class of lattices, and study its connection with the isomorphism class of (V, φ). These are nontrivial, and will be treated in §§2.4 and 2.5. We are then able to classify all the genera of 𝔤-valued hermitian forms in terms of matrices (Theorems 2.10 and 2.11). The results can be presented in simpler forms if K is a real or an imaginary quadratic field of odd class number, in which case the above definitions cover all hermitian spaces. Let d be the discriminant of such a K; then K = Q(\sqrt{d}). For a semi-integral Φ with entries in K, put s(Φ) = p − q when d < 0 and Φ as a complex hermitian matrix has p positive and q negative eigenvalues; we do not define s(Φ) if d > 0. Let ℌ_{n}^{0} be the set of all reduced semi-integral elements of ℌ_{n}. Then we can prove (Theorem 2.14):

(A) Let three integers n, σ, and e be given as follows: 0 < n ∈ 2Z, σ ∈ 2Z, |σ| ≤ n; σ = 0 if d > 0; e is positive and squarefree. Let r be the number of prime factors of e. Suppose that σ−2r ∈ 4Z and no prime factor of e splits in K. Then there exists an element Φ of ℌ_{n}^{0} such that

det(\sqrt{d} Φ) = (−1)^{σ/2}e and s(Φ) = σ if d < 0,
det(\sqrt{d} Φ) = τe with τ = 1 or −1 if d > 0.

Moreover, every element of ℌ_{n}^{0} is of this type. Its genus is determined by (σ, e) if d < 0, and by e if d > 0. If d > 0 and −1 ∈ N_{K/Q}(K^{×}), then both e and −e can occur as det(\sqrt{d} Φ) for Φ in the same genus. If d > 0 and −1 ∉ N_{K/Q}(K^{×}), then τ is uniquely determined by the condition that a prime number p divides e if and only if τe ∉ N_{K/Q}(K_{p}^{×}), where K_{p} = K ⊗_{Q} Q_{p}.

This concerns the case of even n. We have similar but somewhat different results for odd n (Theorem 2.15). In fact we discussed in [S5] and [S6] semiintegral and reduced quadratic forms and obtained results of the same type. If K is imaginary, the hermitian case is almost parallel to the case of quadratic forms over Q, but the theory for real K is more subtle, as can be seen from the above statement.

Let us now turn to the second problem. Before explaining the principal results, let us first discuss natural problems which are more basic and which must be settled before investigating the main question. Given (V, φ) as before, let U^{φ}(V) and SU^{φ}(V) denote the unitary group and the special unitary group of φ, defined as subgroups of GL(V, K). Take an 𝔯-lattice L in V and put

(0.2) Γ (L) = {α ∈ U^{φ}(V) | Lα = L}, Γ^{1}(L) = Γ (L) ∩ SU^{φ}(V).

Then we ask, for a fixed q ∈ F^{×}, whether the set {h ∈ V | φ[h] = q} modulo Γ^{1}(L) is a finite set. A similar question can be asked by replacing F, K, (V, φ) and L by their localizations at a nonarchimedean prime and by defining an obvious analogue of Γ^{1}(L). We will prove that the answer is affirmative in both global and local cases, provided n > 1 (Theorems 3.3 and 4.2). The same is true for the problem about the solutions ξ of the equation ξΦ ·^{t}ξ^{ρ} = Ψ, where Ψ is of size m, and ξ belongs to an 𝔯-lattice in the space of (m × n)-matrices with entries in K, where m is a positive integer < n (Theorems 5.2 and 5.3). We already proved in [S3] the analogues of these facts for quadratic forms and orthogonal groups.

In order to go beyond the mere finiteness, we consider the adelizations U^{φ}(V)_{A} and SU^{φ}(V)_{A}, and define their open subgroups C and C^{1} by

(0.3) C = {γ ∈ U^{φ}(V)_{A} | Lγ = L}, C^{1} = C ∩ SU^{φ}(V)_{A},

where L is a fixed 𝔯-lattice in V. Given two solutions ξ_{0} and ξ_{1} of the equation ξΦ ·^{t}ξ^{ρ} = Ψ, we say that they belong to the same genus (with respect to C) if ξ_{0}γ_{v} = ξ_{1} for every nonarchimedean prime v with an element (γ_{v})_{v} ∈ C. Naturally they are said to belong to the same class if ξ_{0}γ = ξ_{1} with γ ∈ Γ (L). Now to explain our principal ideas in the simplest case, put G = U^{φ}(V) and H = {α ∈ G | ξ_{0}α = ξ_{0}}; also assume for the moment that G_{A} = GC. Then there is a bijection of H\H_{A}/(H_{A} ∩ C) onto the set of classes in the genus of ξ_{0}, and so

(B) #{H\H_{A}/(H_{A} ∩ C)} = the number of classes in the genus of ξ_{0}.

Here and henceforth #{X} denotes the number of elements in a set X. If G_{A} ≠ GC, the right-hand side becomes a finite sum of the class numbers of several genera (Theorem 5.4). Since the left-hand side is the class number of the unitary group H with respect to H_{A} ∩ C, equality (B) connects it to the solutions ξ of ξΦ ·^{t}ξ^{ρ} = Ψ.

If m = 1, the results can be stated in a more transparent way. Returning to the hermitian form φ : V × V → K, put φ[h] = φ(h, h) for h ∈ V. Then the equation ξΦ ·^{t}ξ^{ρ} = Ψ can be written φ[h] = q with h ∈ V and q ∈ F^{×}; thus h and q replace ξ and Ψ. Given a fractional ideal 𝔟 in K, put

(0.4) L[q, 𝔟] = {h ∈ V | φ[h] = q, φ(h, L) = 𝔟}.

We call L integral if φ[x] ∈ 𝔤 for every x ∈ L and call L maximal if it is maximal among the integral 𝔯-lattices. The point of considering L[q, 𝔟] is that L[q, 𝔟], if nonempty, consists of a single genus with respect to C in the above sense. This is clearly a result of local nature; unfortunately, its proof given in Section 3 is not short. In this case H = U^{φ}(W), where W is the orthogonal complement of Kh in V. Now we can prove (Theorem 4.4, Corollary 5.8):

(C) For every y ∈ G_{A}, there is a bijection of H\(H_{A} ∩GyC)/(H_{A} ∩C) onto (Ly^{−1})[q, 𝔟]/Δ_{y}, where Δ_{y} = G ∩ yCy^{−1}.

(D) Take {y_{i}}_{i∈I} ⊂ G_{A} so that G_{A} = ⨆_{i∈I} Gy_{i}C and put L_{i} = Ly_{i}^{−1}. Then

∑_{i∈I} #{L_{i}[q, 𝔟]/Γ (L_{i})} = #{H\H_{A}/(H_{A} ∩ C)}.

(E) If the archimedean factor of G_{A} is compact, then

∑_{i∈I} #{L_{i}[q, 𝔟]}/#{Γ_{i}} = 𝔪(H, H_{A} ∩ C),

where the right-hand side is the mass of H with respect to the open subgroup H_{A} ∩ C of H_{A} in the sense of [S2].

These assertions are true with SU^{φ}(V), SU^{φ}(W) and C^{1} in place of G, H, and C, if dim(V) > 2 and we impose a certain condition on (q, 𝔟). Notice that (E) gives the mass of H by means of the number of solutions h of φ[h] = q under the condition φ(h, L) = 𝔟, while (D) gives the class number of H.

In our recent book [S3, Chapter III] we developed a theory of a Diophantine equation φ[h] = q for a quadratic form φ defined on a vector space over an algebraic number field. The principal result is that to each “primitive solution” h of this equation for a fixed q, considered modulo the group of units Γ, one can associate a “class” of lattices with respect to the orthogonal group H of the restriction of φ to a subspace of codimension 1. Consequently the class number of H equals the number of such h modulo Γ. This includes as a special case the result of Gauss that the number of primitive representations of q as the sum of three squares equals an elementary factor times the class number of primitive binary quadratic forms of discriminant −q. Also, formulas of type (B) and (E) were proved in [S3] and [S6] for quadratic forms. The reader is referred to [S7] for some more historical and technical comments on this subject. Now (B), (C), (D), and (E) are hermitian analogues of these results. In order to develop the theory for hermitian forms, we are naturally guided by the formulation in the case of quadratic forms, but we need new ideas and technique, and it is wrong to say that we can do things “in the same way.” This is especially so when we consider the problem with respect to the special unitary group instead of the unitary group. Thus there are two theories with respect to these two types of groups, and one, that for the special unitary group, is more complex than the other, and in a sense more interesting.

1. Generalities on hermitian forms and unitary groups

1.1. For an associative ring A with an identity element we denote by A^{×} the group of all invertible elements of A, and for positive integers m and n we denote by A_{n}^{m} the A-module of all (m × n)-matrices with entries in A. We put M_{n}(A) = A_{n}^{n} when we view it as a ring, and denote by 1_{n} its identity element.

We take a basic field F and a couple (K, ρ) consisting of an F -algebra K and a nontrivial F -linear automorphism ρ of K belonging to the following two types:

(I) K is a separable quadratic extension of F and ρ generates Gal(K/F); (II) K = F × F and (x, y)^{ρ} = (y, x) for (x, y) ∈ F × F.

In our later discussion, K of type (II) will appear as a localization of a global K of type (I). For a matrix α = (a_{ij}) with entries in K we denote by ^{t}α the transpose of α, and put α^{ρ} = (a_{ij}^{ρ}) and α^{∗} = ^{t}α^{ρ}; we put also α^{−ρ} = (α^{ρ})^{−1} when α is invertible. For a subring S of K we write α ≺ S if all the entries of α are contained in S. Given a left K-module V, we denote by End(V, K) the ring of all K-linear endomorphisms of V and put GL(V, K) = End(V, K)^{×}. We let End(V, K) act on V on the right; namely we denote by wα the image of w ∈ V under α ∈ End(V, K).

Let V be a left K-module isomorphic to K_{n}^{1}; we put then n = dim(V). By a hermitian space we mean a structure (V, φ), where φ is a hermitian form on V, that is, an F -bilinear map φ : V × V → K such that

(1.1) φ(x, y)^{ρ} = φ(y, x), (1.2) φ(ax, by) = ab^{ρ}φ(x, y) for every a, b ∈ K.

Whenever we speak of a hermitian space (V, φ), we assume that φ is nondegenerate, and put φ[x] = φ(x, x) for x ∈ V. We define groups U^{φ}(V) and SU^{φ}(V) by

(1.3a) U^{φ} = U^{φ}(V) = {α ∈ GL(V, K) | φ[xα] = φ[x] for every x ∈ V }, (1.3b) SU^{φ} = SU^{φ}(V) = {α ∈ U^{φ}(V) | det(α) = 1}.

For every free K-submodule X of V on which φ is nondegenerate, we put

(1.4) X^{⊥} = {y ∈ V | φ(y, X) = 0},

and define U^{φ}(X) and SU^{φ}(X) by (1.3a, b) with X in place of V; namely we use φ for its restriction to X. We always identify U^{φ}(X) with the subgroup of U^{φ}(V) consisting of the elements α such that yα = y for every y ∈ X^{⊥}. Similarly we view SU^{φ}(X) as a subgroup of SU^{φ}(V).

Let h be an element of V such that φ[h] ≠ 0. Then

(1.5) \{x ∈ V | φ[x] = φ[h]\} = \{ \begin{matrix}h · U^{φ} if dim(V) = 1, \\ h · SU^{φ} if dim(V) > 1. \end{matrix}

This follows easily from the generalized Witt theorem; see [S2, Lemma 1.3], for example. The case K = F × F is not included in that theorem, but the structure of (V, φ) for such a K is determined by dim(V) as shown in [S2, §2.13], and so the fact corresponding to the Witt theorem is trivially true; see also §1.8 below.

1.2. Let φ_{0} be the matrix that represents φ with respect to a K-basis of V; then we denote by d_{0}(V, φ) the element of F^{×}/N_{K/F} (K^{×}) represented by (−1)^{n/2} det(φ_{0}) or (−1)^{(n−1)/2} det(φ_{0}) according as n is even or odd. This does not depend on the choice of a K-basis of V. We denote d_{0}(V, φ) simply by d_{0}(V) or d_{0}(φ) when there is no fear of confusion.

Given s ∈ F^{×}, we denote by {K, s} the quaternion algebra B over F in which K is embedded and which is given by

(1.6) B = K + Kω, ω^{2} = s, ωa = a^{ρ}ω for every a ∈ K.

Since B is determined by K and sN_{K/F} (K^{×}), for a coset ε ∈ F^{×}/N_{K/F} (K^{×}) we denote by {K, ε} the algebra {K, s} with any s ∈ ε. In particular, we can associate with (V, φ) a quaternion algebra {K, d_{0}(φ)}.

Lemma 1.3. Given (V, φ) as in §1.1, suppose dim(V) = 2 and put B = {K, d_{0}(φ)}. Then there is a ring-injection j of B into End(V, K) and an element ℓ of V such that φ[ℓ] ≠ 0, ℓj(B) = V, ℓj(a) = aℓ for every a ∈ K, and φ[ℓj(ξ)] = φ[ℓ]ξξ^{ι} for every ξ ∈ B, where ι is the main involution of B. Moreover, Tr_{K/F} (φ(ℓα, ℓβ)) = φ[ℓ]Tr_{B/F} (αβ^{ι}) for every α, β ∈ B, and

(1.7a) U^{φ}(V) = {z^{−1}α | z ∈ K^{×}, α ∈ B^{×}, zz^{ρ} = αα^{ι}}, (1.7b) SU^{φ}(V) = {α ∈ B^{×} | αα^{ι} = 1},

where we identify α with j(α) for α ∈ B.

Proof. Identify V with K_{2}^{1} so that φ(x, y) = xφ_{0}y^{∗} for x, y ∈ K_{2}^{1} with φ_{0} = diag[c, −cs], where c, s ∈ F^{×}. Then d_{0}(φ) = sN_{K/F} (K^{×}), and so B is

given by (1.6). Define j : B → M_{2}(K) by j(a + bω) = [ \begin{matrix}a & b \\ sb^{ρ} & a^{ρ} \end{matrix} ] and put ℓ =

(1, 0). Then it is an easy exercise to verify all the statements; cf. [S2, Lemmas 4.3, 4.4, and (4.3.2)]. Notice that j(B) = {α ∈ M_{2}(K) | α^{ι}φ_{0} = φ_{0}α^{∗}}.

1.4. When K is a field, by a weak Witt decomposition of V we mean a direct sum decomposition of V with 2r elements e_{i}, f_{i}, and a subspace Z of V such that

(1.8a)  V = ∑_{i=1}^{r}(Ke_{i} + Kf_{i}) + Z,  Z = ( ∑_{i=1}^{r}(Ke_{i} + Kf_{i}))^{⊥},

(1.8b) φ(e_{i}, e_{j}) = φ(f_{i}, f_{j}) = 0, φ(e_{i}, f_{j}) = δ_{ij} for every i and j.

Clearly ∑_{i=1}^{r}(Ke_{i} + Kf_{i}) is a subspace of dimension 2r. We call this a Witt decomposition if φ[x] ≠ 0 for every x ∈ Z, ≠ 0, in which case we call Z a core subspace of (V, φ) and dim(Z) the core dimension of (V, φ). If ζ is the restriction of φ to Z, then clearly d_{0}(φ) = d_{0}(ζ).

1.5. In this paper a global field means an algebraic number field of finite degree, and a local field the completion of a global field at a nonarchimedean prime. For a global field F we denote by 𝔤 the ring of algebraic integers in F; for a local F we denote by 𝔤 the ring of local integers in F in the standard sense. An archimedean completion of a global field will not be called a local field. In both local and global cases, by a 𝔤-lattice in a finite-dimensional vector space V over F, we mean a finitely generated 𝔤-submodule of V that spans V over F.

Let (K, ρ) be as in §1.1 with a local or global F. We then denote by 𝔯 the ring of all elements of K integral over 𝔤, and by 𝔡 the different of K relative to F. We have 𝔯 = 𝔡 = 𝔤 × 𝔤 if K = F × F. By a 𝔤-ideal we mean a fractional ideal in F, and similarly by an 𝔯-ideal we mean a fractional ideal in K if K is a field. If K = F × F, an 𝔯-ideal means a subset of K of the form 𝔞 × 𝔟 with 𝔤-ideals 𝔞 and 𝔟. In both local and global cases, by an 𝔯-lattice in a K-module V as in §1.1 we mean a 𝔤-lattice in V stable under multiplication by the elements of 𝔯. Given two 𝔯-lattices L and M in V, we denote by [L/M] the 𝔯-ideal generated by det(α) for all α ∈ GL(V) such that Lα ⊂ M. Thus [L/Lα] = det(α)𝔯. In particular, if α ∈ U^{φ} and K is a field in the local case, then [L/Lα] = 𝔯. If K = F × F, however, [L/Lα] = 𝔞 × 𝔞^{−1} with a 𝔤-ideal 𝔞 for α ∈ U^{φ}.

Lemma 1.6. Two hermitian spaces (V, φ) and (V′, φ′) in the local case are isomorphic if and only if dim(V) = dim(V′) and d_{0}(φ) = d_{0}(φ′).

This is well known. For the proof, see [S2, Proposition 5.3], for example.

1.7. Let (V, φ) be defined with a local or global F. For a 𝔤-lattice L in V we denote by μ(L) the 𝔤-ideal generated by φ[x] for all x ∈ L. We call a 𝔤-lattice L in V integral if μ(L) ⊂ 𝔤; we call an 𝔯-lattice L maximal if L is maximal among the integral 𝔯-lattices. (This is what we called 𝔤-maximal in [S2].) For basic properties of maximal lattices in V the reader is referred to [S1] or Sections 4 and 5 of [S2]. For example, we note ([S2, (4.7.1)])

(1.9) φ(L, L) ⊂ 𝔡^{−1} if L is integral.

If n > 1 and L is maximal, then μ(L) = 𝔤. This fact in the global case follows from the local case, which can be seen from [S2, Lemmas 5.4 and 5.6].

Given an 𝔯-lattice L in V, q ∈ F^{×}, and an 𝔯-ideal 𝔟, we put

(1.10a)  L̂ = {x ∈ V | φ(x, L) ⊂ 𝔡^{−1}},

(1.10b) L[q] = {x ∈ L | φ[x] = q}, (1.10c) L[q, 𝔟] = {x ∈ V | φ[x] = q, φ(x, L) = 𝔟}.

By (1.9), we have L ⊂ L̂ if L is integral. The set L[q, 𝔟] is not necessarily contained in L[q]. If M is another 𝔯-lattice in V, then we easily see that [L/M]^{ρ} = [M̂/L̂]. If L_{1} = Lα with α ∈ U^{φ}, then L̂_{1} = L̂α, and so [L̂_{1}/L_{1}] = [L̂/L].

The notation being as in (1.8a, b), take a maximal 𝔯-lattice M in Z and put (1.11) L = ∑_{i=1}^{r}(𝔯e_{i} + 𝔡^{−1}f_{i}) + M.

Then L is maximal; see [S2, Lemma 4.9 (2)]. We can easily verify that

(1.12)  L̂ = ∑_{i=1}^{r}(𝔯e_{i} + 𝔡^{−1}f_{i}) + M̂, M̂ = {x ∈ Z | φ(x, M) ⊂ 𝔡^{−1}}.

1.8. Let us now consider the case K = F × F; then we define the core dimension of (V, φ) to be 0. We can write an element of V = K_{n}^{1} in the form (x, y) with x, y ∈ F_{n}^{1}. Taking a suitable coordinate system, we can assume that

(1.13)  φ((x, y), (z, w)) = (x ·^{t}w, y ·^{t}z)  (x, y, z, w ∈ F_{n}^{1}).

This is shown in [S2, §2.13]. We have then φ[(x, y)] = x ·^{t}y and

(1.14) U^{φ} = {(ξ, ξ̃) | ξ ∈ GL_{n}(F)}, SU^{φ} = {(ξ, ξ̃) | ξ ∈ SL_{n}(F)},

where ξ̃ = ^{t}ξ^{−1}; also 𝔤_{n}^{1} × 𝔤_{n}^{1} is a maximal lattice. It should be noted that if h ∈ V and φ[h] ≠ 0, then Kh is isomorphic to K.

Lemma 1.9. Let L be a maximal lattice in V and t the core dimension of (V, φ); suppose K is a local field; put

E = {e ∈ 𝔯^{×} | ee^{ρ} = 1}, E_{0} = {e^{ρ}/e | e ∈ 𝔯^{×}}, E_{L} = det (C(L)),

where C(L) = {α ∈ U^{φ}(V) | Lα = L}. Then the following assertions hold: (i) [E : E_{0}] = 1 or 2 according as K is unramified or ramified over F. (ii) E_{L} = E_{0} if t = 0; E_{L} = E if t > 0.

This is included in [S1, Lemma 4.16 and Proposition 4.18] and [S2, Lemma 5.11].

Suppose K is a local field ramified over F; let L and M be maximal lattices in V. Then there exists an element α ∈ U^{φ} such that M = Lα as shown in (i) of the following lemma. We then denote by e(L/M) the element of E/E_{0} represented by det(α). This is well defined in view of (ii) above.

Lemma 1.10. Let L and M be maximal lattices in V in the local case, and let t be the core dimension of (V, φ). Then the following assertions hold:

(i) There exists an element α of U^{φ} such that M = Lα.

(ii) We can take such an α from SU^{φ} if t > 0 or K is a field unramified over F.

(iii) Suppose K is ramified over F and t = 0; then M = Lα with α ∈ SU^{φ} if and only if e(L/M) = 1.

(iv) Suppose K = F × F; then M = Lα with α ∈ SU^{φ} if and only if [L/M] = 𝔯.

Proof. The first assertion is included in [S1, Propositions 3.3 and 4.13] and also in [S2, Lemmas 4.12 and 5.9]. Next, suppose K is a field; given M, take α ∈ U^{φ} so that Lα = M and put c = det(α). Then c ∈ E. If K is unramified over F or t > 0, then Lemma 1.9 guarantees an element β of U^{φ} such that Lβ = L and det(β) = c^{−1}. Then βα ∈ SU^{φ} and Lβα = M. This proves (ii). Assertion (iii) can be proved in a similar way. Assertion (iv) is included in [S1, Proposition 3.3].

1.11. Let us now consider the case with n = 2 and a local F. Using the symbols B, j, and ℓ of Lemma 1.3, we first observe that

(1.15) B ≅ M_{2}(F) ⟺ t = 0 ⟺ 1 ∈ d_{0}(φ).

Let 𝔒 be a maximal order in B containing 𝔯. We can find an element γ ∈ B^{×} such that φ[ℓ]^{−1} = γγ^{ι}. Put M = ℓj(𝔒γ) and 𝔒̂ = {α ∈ B | Tr_{B/F} (α𝔒) ⊂ 𝔤}. Identifying B with j(B), for α ∈ B we see that

ℓαγ ∈ M̂ ⟺ φ(ℓαγ, M) ⊂ 𝔡^{−1} ⟺ Tr_{K/F} (𝔯φ(ℓαγ, M)) ⊂ 𝔤
⟺ Tr_{B/F} (α𝔒) ⊂ 𝔤 ⟺ α ∈ 𝔒̂.

Thus M̂ = ℓj(𝔒̂γ). If B is not a division algebra, then 𝔒̂ = 𝔒, so that M̂ = M, which means that M is maximal. Suppose B is a division algebra; then 𝔒 = {α ∈ B | αα^{ι} ∈ 𝔤} as noted in [S3, Theorem 5.13], and so M = {x ∈ V | φ[x] ∈ 𝔤}, which is a unique maximal lattice in V by [S2, Lemma 5.4]. Since 𝔒̂ = 𝔓^{−1} with the maximal ideal 𝔓 of 𝔒, we have M̂ = ℓj(𝔓^{−1}γ). Thus [M̂/M] = 𝔭𝔯 with the maximal ideal 𝔭 of 𝔤.

2. Classification of hermitian forms over a global field

2.1. Throughout this section we assume that F is a global field and K is a quadratic extension of F. We denote by a and h the sets of archimedean primes and nonarchimedean primes of F respectively, and put v = a ∪ h. Given an algebraic group G defined over F, we define G_{v} for each v ∈ v and the adelization G_{A} as usual, and view G and G_{v} as subgroups of G_{A}. We then denote by G_{a} and G_{h} the archimedean and nonarchimedean factors of G_{A}, respectively. In particular, the adelization of the multiplicative group F^{×} is denoted by F_{A}^{×}, which is the idele group of F. For x ∈ G_{A} and v ∈ v we denote by x_{v} the v-component of x.

Given (V, φ) over F, for each v ∈ v we can define the v-localization (V, φ)_{v} = (V_{v}, φ_{v}) with φ_{v} : V_{v} × V_{v} → K_{v} in a natural way. For v ∈ h let t_{v} be the core dimension of (V, φ)_{v}. Since x ↦ φ_{v}[x] for x ∈ V_{v} can be viewed as an F_{v}-valued quadratic form, we have 2t_{v} ≤ 4 by a well known principle, and so t_{v} ≤ 2. Let r_{0} denote the set of all real archimedean primes of F that do not split in K. If v ∈ a and v ∉ r_{0}, then there is only one isomorphism class of (V, φ)_{v} for each n. For each fixed v ∈ r_{0} we have a pair of nonnegative integers (p_{v}, q_{v}) such that φ_{v} is represented by diag[1_{p_{v}}, −1_{q_{v}}] when F_{v} and K_{v} are identified with R and C. We put then s_{v}(φ) = p_{v} − q_{v}, and call s_{v}(φ) the index of φ at v. Clearly |s_{v}(φ)| is the core dimension of φ_{v}, s_{v}(φ) − n ∈ 2Z, and |s_{v}(φ)| ≤ n; also, n and s_{v}(φ) determine (p_{v}, q_{v}), and vice versa.

For an 𝔯-lattice L in V and v ∈ h we denote by L_{v} the 𝔯_{v}-linear span of L in V_{v}. Also, for ξ ∈ GL(V, K)_{A} we denote by Lξ the lattice in V such that (Lξ)_{v} = L_{v}ξ_{v} for every v ∈ h. By the U^{φ}(V)-genus (resp. SU^{φ}(V)-genus) of L we understand the set of all lattices of the form Lξ with ξ ∈ U^{φ}(V)_{A} (resp. ξ ∈ SU^{φ}(V)_{A}). Also, by the U^{φ}(V)-class (resp. SU^{φ}(V)-class) of L we understand the set of all lattices of the form Lα with α ∈ U^{φ}(V) (resp. α ∈ SU^{φ}(V)).

The classification of (V, φ) over a number field was done by Landherr in [L]. We formulate the results in the form that suits our later purposes, and give a proof for the reader’s convenience. To be precise, we are going to show that the isomorphism classes of hermitian spaces correspond bijectively to the sets of data consisting of the following objects:

(2.1a) 0 < n ∈ Z; ε ∈ F^{×}; an integer σ_{v}, given for each v ∈ r_{0}, such that |σ_{v}| ≤ n and σ_{v} − n ∈ 2Z.

We look for (V, φ) such that dim(V) = n, d_{0}(φ) is represented by ε, and s_{v}(φ) = σ_{v} for every v ∈ r_{0}. Clearly the following condition is necessary:

(2.1b) (−1)^{σ_{v}/2}ε > 0 for every v ∈ r_{0} if n ∈ 2Z and (−1)^{(σ_{v}−1)/2}ε > 0 for every v ∈ r_{0} if n − 1 ∈ 2Z.

Theorem 2.2. (i) The isomorphism class of (V, φ) is determined by n, {σ_{v}}, and d_{0}(φ).

(ii) Given n, ε, and {σ_{v}} satisfying (2.1a, b), there exists a hermitian space (V, φ) such that dim(V) = n, ε ∈ d_{0}(φ), and s_{v}(φ) = σ_{v} for every v ∈ r_{0}.

Proof. Clearly n and {σ_{v}} determine (V, φ)_{v} for every v ∈ a, and n and d_{0}(φ) determine (V, φ)_{v} for every v ∈ h by Lemma 1.6. Therefore we obtain (i) in view of the Hasse principle. We prove (ii) by induction on n. The case n = 1 is trivial, and so we assume n > 1. We first prove the case in which σ_{v} ≥ 0 for every v ∈ r_{0}. Let τ_{v} = σ_{v} − 1. Then the set (n − 1, (−1)^{n−1}ε, {τ_{v}}) satisfies (2.1a, b), and therefore by induction we can find a hermitian space (W, ψ) such that dim(W) = n − 1, τ_{v} = s_{v}(ψ) for every v ∈ r_{0}, and (−1)^{n−1}ε ∈ d_{0}(ψ). Put V = K ⊕ W and define φ on V by φ[a ⊕ y] = aa^{ρ} + ψ[y] for a ∈ K and y ∈ W. Then clearly ε ∈ d_{0}(φ) and s_{v}(φ) = σ_{v} for every v ∈ r_{0}.

Now, given {σ_{v}} with possibly negative σ_{v}, take c ∈ F^{×} so that c < 0 or c > 0 at v ∈ r_{0} according as σ_{v} < 0 or σ_{v} ≥ 0. Then the set (n, c^{n}ε, {|σ_{v}|}) satisfies (2.1a, b). Therefore we can find a hermitian space (V_{1}, φ_{1}) such that dim(V_{1}) = n, c^{n}ε ∈ d_{0}(φ_{1}), and s_{v}(φ_{1}) = |σ_{v}| for every v ∈ r_{0}. Put φ = cφ_{1}. Then ε ∈ d_{0}(φ) and s_{v}(φ) = σ_{v} for every v ∈ r_{0}. This completes the proof.

Theorem 2.3. Given (V, φ), put B = {K, d_{0}(φ)} using the notation of §1.2. Let 𝔢 be the product of the prime ideals of F ramified in B; also let L be a maximal lattice in V. Then the following assertions hold.

(i) [L̂/L] = 𝔢𝔯 if n is even.

(ii) When n is odd, put d_{0}(φ)𝔤 = 𝔞N_{K/F} (𝔟) with an 𝔯-ideal 𝔟 and a squarefree integral 𝔤-ideal 𝔞 whose prime factors remain prime in K. Then [L̂/L] = 𝔞𝔡, where 𝔡 is the different of K relative to F.

Proof. For v ∈ h let t_{v} be the core dimension of (V, φ)_{v}. Suppose n ∈ 2Z; then t_{v} = 0 if and only if d_{0}(φ) is represented by an element of N_{K/F} (K_{v}^{×}), that is, if and only if v does not divide 𝔢. If n is odd, the isomorphism class of (V, φ)_{v} depends on 𝔞_{v} and 𝔡_{v}. Thus our assertions can be reduced to the question about [L̂_{v}/L_{v}] for v ∈ h. In fact, suppressing the subscript v, we have, in the local case,

(2.2) [L̂/L] = 𝔯 if t = 0; [L̂/L] = 𝔭𝔯 if t = 2; [L̂/L] = 𝔡 if t = 1 and d_{0}(φ) ∩ 𝔤^{×} ≠ ∅; [L̂/L] = 𝔭𝔡 if t = 1 and d_{0}(φ) ∩ 𝔤^{×} = ∅.

Here 𝔭 is the maximal ideal of 𝔤. In view of Lemma 1.10 (i), it is sufficient to prove this for a special choice of L. If K = F × F, then we can put L = 𝔤_{n}^{1} × 𝔤_{n}^{1} as noted in §1.8, and so L̂ = L. Thus we assume that K is a field. By (1.12), L = L̂ if t = 0. Let M = {x ∈ Z | φ[x] ∈ 𝔤}. By (1.12), [L̂/L] = [M̂/M]. We have seen that [M̂/M] = 𝔭𝔯 in §1.11 if t = 2. If t = 1, then M = 𝔯ℓ with an element ℓ such that φ[ℓ]𝔤 is 𝔤 or 𝔭𝔤. Thus M̂ = 𝔡^{−1}φ[ℓ]^{−1}ℓ, and so [L̂/L] = [M̂/M] = φ[ℓ]𝔡, which completes the proof of (2.2). Combining the results on [L̂_{v}/L_{v}] for all v ∈ h, we obtain our theorem.

2.4. To illustrate Theorem 2.3 in terms of matrices, we have to define the genus and class of a hermitian matrix. We put

(2.3)  𝔊 = GL_{n}(K),  ℌ_{n} = {Φ ∈ 𝔊 | Φ^{∗} = Φ},  L_{0} = 𝔯_{n}^{1},
(2.4)  E = 𝔊_{a} ∏_{v∈h} GL_{n}(𝔯_{v}),  E_{ξ} = ξ^{−1}Eξ (ξ ∈ 𝔊_{A}),
(2.5)  Δ_{ξ} = E_{ξ} ∩ 𝔊,  Δ_{ξ}^{1} = E_{ξ} ∩ SL_{n}(K).

Every 𝔯-lattice L in K_{n}^{1} can be given as L = L_{0}ξ with ξ ∈ 𝔊_{A}, and E_{ξ} = {y ∈ 𝔊_{A} | Ly = L}. We denote by ℌ_{n}(ξ) the set of all Φ ∈ ℌ_{n} such that xΦx^{∗} ∈ 𝔤 for every x ∈ L_{0}ξ. We call such a Φ reduced (relative to ξ) if the following condition is satisfied:

(2.6) Φ ∈ ℌ_{n}(ζ^{−1}ξ) with ζ ∈ 𝔊_{h} ∩ ∏_{v∈h} M_{n}(𝔯_{v}) ⟹ ζ ∈ E.

We denote by ℌ_{n}^{0}(ξ) the set of all reduced elements of ℌ_{n}(ξ).

We say that two elements Φ and Ψ of ℌ_{n}(ξ) belong to the same genus (relative to ξ) if there exists an element ε of E_{ξ} such that εΦε^{∗} = Ψ; they are said to belong to the same U-class (resp. SU-class) if αΦα^{∗} = Ψ for some α ∈ Δ_{ξ} (resp. α ∈ Δ_{ξ}^{1}). These depend on the choice of L = L_{0}ξ.

Given Φ ∈ ℌ_{n}, put V = K_{n}^{1} and φ[x] = xΦx^{∗} for x ∈ V. Then we obtain a hermitian space (V, φ), which we denote by [Φ], and we write U(Φ) and SU(Φ) for U^{φ}(V) and SU^{φ}(V) as subgroups of 𝔊. Put L = L_{0}ξ with ξ ∈ 𝔊_{A}. Clearly L is integral if Φ ∈ ℌ_{n}(ξ), in which case L is maximal if and only if Φ ∈ ℌ_{n}^{0}(ξ). Thus an element of ℌ_{n}^{0}(ξ) determines a hermitian space and a maximal lattice.

2.5. To parametrize all genera of Φ ∈ ℌ_{n}^{0}(ξ), we need a few more symbols:

(2.7)  𝔱 = K_{A}^{×} ∩ (K_{a}^{×} ∏_{v∈h} 𝔱_{v}),  𝔱_{v} = {y ∈ K_{v}^{×} | yy^{ρ} ∈ 𝔤_{v}^{×}},

(2.8) T = {x ∈ 𝔊_{A} | det(x) ∈ 𝔱}. Notice that 𝔱_{v} = 𝔯_{v}^{×}·{y ∈ K_{v}^{×} | yy^{ρ} = 1} for every v ∈ h, and 𝔱_{v} = 𝔯_{v}^{×} if v does not split in K. Let ℑ_{K} denote the ideal group of K and ℑ_{K/F}^{0} the subgroup of ℑ_{K} generated by the ideals 𝔞 such that N_{K/F} (𝔞) = 𝔤 and the principal ideals. Now there is a sequence of isomorphisms:

(2.9) 𝔊_{A}/T 𝔊 ≅ K_{A}^{×}/K^{×}𝔱 ≅ ℑ_{K}/ℑ_{K/F}^{0} .

The last isomorphism can be obtained by the map y ↦ y𝔯 for y ∈ K_{A}^{×}. As for the first isomorphism, we first note, for every ξ ∈ 𝔊_{A} and Φ ∈ ℌ_{n},

(2.10) T 𝔊ξ = EξU(Φ)_{A}𝔊 = {x ∈ 𝔊_{A} | det(ξ^{−1}x) ∈ K^{×}𝔱}.

Clearly the last set contains the second set. Conversely, suppose x ∈ 𝔊_{A} and det(ξ^{−1}x) = by with b ∈ K^{×} and y ∈ 𝔱. We can find z, w ∈ K_{A}^{×} such that z_{v} ∈ 𝔯_{v}^{×} and w_{v}w_{v}^{ρ} = 1 for every v ∈ h and y = zw. We can find ε ∈ E, α ∈ 𝔊, and γ ∈ U(Φ)_{A} such that det(ε) = z, det(α) = b, and det(γ) = w. Then det(x^{−1}εξγα) = 1. By strong approximation in SL_{n}(K) we see that x^{−1}εξγα ∈ x^{−1}ExSL_{n}(K), and so x^{−1}εξγα = x^{−1}ε′xβ with ε′ ∈ E and β ∈ SL_{n}(K). Then x = (ε′)^{−1}εξγαβ^{−1} ∈ EξU(Φ)_{A}𝔊, which proves the last equality of (2.10). That T 𝔊 equals the last set of (2.10) for ξ = 1 can be proved in the same way. Thus T 𝔊 is the inverse image of K^{×}𝔱 under the map x ↦ det(x), and so T 𝔊 is a normal subgroup of 𝔊_{A}. Then we obtain the first isomorphism of (2.9) and also the first equality of (2.10) for every ξ ∈ 𝔊_{A}.

Proposition 2.6. (i) For Φ, Ψ ∈ ℌ_{n}^{0}(ξ), ξ ∈ 𝔊_{A}, the spaces [Φ] and [Ψ] are isomorphic if and only if they belong to the same genus.

(ii) Let X be a complete set of representatives for 𝔊_{A}/T 𝔊, and for each ξ ∈ 𝔊_{A} let Y_{ξ} be a complete set of representatives for the genera of the elements of ℌ_{n}^{0}(ξ). Then the hermitian spaces [Φ] obtained from Φ ∈ Y_{ξ} for all ξ ∈ X exhaust all isomorphism classes of n-dimensional hermitian spaces without overlapping.

Proof. Let Φ and Ψ be elements of ℌ_{n}^{0}(ξ) belonging to the same genus. Then there exists an element ε ∈ E_{ξ} such that εΦε^{∗} = Ψ, and the Hasse principle guarantees an element α of 𝔊 such that Ψ = αΦα^{∗}. Thus [Ψ] is isomorphic to [Φ]. Conversely, suppose [Φ] and [Ψ] are isomorphic for Φ, Ψ ∈ ℌ_{n}^{0}(ξ). Then Φ = βΨβ^{∗} for some β ∈ 𝔊. Now L_{0}ξ is maximal in both [Φ] and [Ψ], and L_{0}ξβ is maximal in [Ψ]. Thus L_{0}ξβ = L_{0}ξγ with γ ∈ U(Ψ)_{A} by Lemma 1.10. Put ζ = βγ^{−1}. Then ζ ∈ E_{ξ}, and ζΨζ^{∗} = Φ. Therefore Ψ belongs to the genus of Φ. This proves (i). Clearly every n-dimensional hermitian space is isomorphic to [Ψ] for some Ψ ∈ ℌ_{n}. Take a maximal lattice L in K_{n}^{1} = V and put L = L_{0}η with η ∈ 𝔊_{A}. We have then η ∈ T 𝔊ξ with some ξ ∈ X. By (2.10) we can put ξ = εηγα^{−1} with ε ∈ E, γ ∈ U(Ψ)_{A}, and α ∈ 𝔊. Put Φ = αΨα^{∗}. Then α gives an isomorphism of [Φ] onto [Ψ]. Now L_{0}ηγ is a maximal lattice in [Ψ] and L_{0}ξα = L_{0}ηγ, and so L_{0}ξ is a maximal lattice in [Φ]. Thus Φ ∈ ℌ_{n}^{0}(ξ). By (i), Φ can be replaced by a member of Y_{ξ}. This shows that every (V, φ) can be obtained as described in (ii). Now suppose [Φ_{1}] and [Φ_{2}] are isomorphic for Φ_{i} ∈ Y_{ξ_{i}} with ξ_{1}, ξ_{2} ∈ X. Then Φ_{1} = αΦ_{2}α^{∗} with α ∈ 𝔊. Since L_{0}ξ_{i} is maximal, we have L_{0}ξ_{1}α = L_{0}ξ_{2}ζ with ζ ∈ U(Φ_{2})_{A}. Then ξ_{1}αζ^{−1}ξ_{2}^{−1} ∈ E, and so det(ξ_{1}ξ_{2}^{−1}) ∈ K^{×}𝔱. We are taking the ξ_{i} from X, and therefore ξ_{1} = ξ_{2} by (2.10). By (i), Φ_{2} belongs to the genus of Φ_{1}, and so Φ_{2} = Φ_{1}. This completes the proof.

2.7. The connection of a class of hermitian matrices with a class of lattices is not so simple in general. Given Φ ∈ ℌ_{n}^{0}(ξ) with a fixed ξ, in order to exhaust all classes in the U(Φ)-genus of L_{0}ξ, we have to consider the genera of elements in ℌ_{n}^{0}(ξζ) for all ζ ∈ T 𝔊/E_{ξ}𝔊. Thus ℌ_{n}^{0}(ξ) is sufficient if and only if K^{×}𝔱 = K^{×} det(E). We will not go into details, as we do not need the result in our later treatment.

The case of SU-class is simpler. Fix ξ ∈ 𝔊 and put L = L_{0}ξ. For Φ ∈ ℌ_{n}(ξ) we define the SU-genus (relative to ξ) of Φ to be the set of all Ψ ∈ ℌ_{n}^{0}(ξ) such that Ψ = εΦε^{∗} with ε ∈ E_{ξ} such that det(ε) = 1. Clearly det(Ψ) = det(Φ). Given such Ψ and ε, the Hasse principle guarantees an element α ∈ 𝔊 such that Ψ = αΦα^{∗}. Then det(α) det(α)^{ρ} = 1. Changing α for αγ with a suitable γ ∈ U(Φ), we may assume that det(α) = 1. Since Lα = Lε^{−1}α and ε^{−1}α ∈ SU(Φ)_{A}, we see that Lα belongs to the SU(Φ)-genus of L. We then associate the SU(Φ)-class of Lα to Ψ. We can easily verify that the set of all SU-classes in the SU-genus of Φ contained in ℌ_{n}(ξ) are in one-to-one correspondence with the set of SU(Φ)-classes in the SU(Φ)-genus of L.

2.8. Define (V, φ) by V = K_{n}^{1} and φ[x] = xΦx^{∗} as above with any Φ ∈ ℌ_{n}. Put L = L_{0}ξ with ξ ∈ 𝔊_{A}. We easily see that L̂ = 𝔡^{−1}L_{0}(Φξ^{∗})^{−1}, and so

(2.11) [L̂/L] = det(Φξξ^{ρ})𝔡^{n} if L = L_{0}ξ.

We need a few more symbols. First, we put 𝔡_{0} = 𝔡^{2} ∩ F. For v ∈ r_{0} and (V, φ) isomorphic to [Φ] with Φ ∈ ℌ_{n} we put s_{v}(Φ) = s_{v}(φ) and d_{0}(Φ) = d_{0}(φ).

Lemma 2.9. Let B be a quaternion algebra over F and K a quadratic extension of F contained in B; let 𝔯 be the maximal order of K and 𝔒 a maximal order in B containing 𝔯; further let 𝔢 be the product of the prime ideals in F ramified in B and 𝔡 the different of K relative to F. Then there exists a 𝔤-ideal 𝔞 such that 𝔒 is isomorphic as a left 𝔯-module to 𝔯⊕𝔞 and N_{K/F} (𝔡𝔞) = s𝔢 with an element s such that B is isomorphic to {K, s}. Moreover, the coset 𝔞ℑ_{K/F}^{0} is independent of the choice of 𝔒, and 𝔒 is isomorphic as a right 𝔯-module to 𝔯 ⊕ 𝔞^{ι}, where ι is the main involution of B.

Proof. Take ε ∈ F^{×} so that B = {K, ε}, and consider (V, φ) = [Φ] over K with V = K_{2}^{1} and φ such that ε ∈ d_{0}(φ). Using the symbols y, j of Lemma 1.3, identify j(α) with α for α ∈ B, and put M = y𝔒γ with γ ∈ B_{h}^{×} such that φ[y]^{−1} = γ_{v}γ_{v}^{ι} for every v ∈ h. Applying the local result of §1.11 to M_{v}, we see that M is maximal and [M̂/M] = 𝔢𝔯. Put M = 𝔯_{2}^{1}ξ with some ξ ∈ GL_{2}(K)_{A} and 𝔞 = det(ξ)𝔯. Then by a well-known principle M is 𝔯-isomorphic to 𝔯 ⊕ 𝔞. By (2.11) we have det(Φ)N_{K/F} (𝔡𝔞) = 𝔢. Then we obtain the first assertion of our lemma by taking s = − det(Φ)^{−1}. Let 𝔒′ be another maximal order in B containing 𝔯. By the Chevalley-Hasse-Noether theorem (see [E, Satz 7]) there exists an 𝔯-ideal 𝔟 such that 𝔟𝔒′ = 𝔒𝔟. Take c ∈ K_{A}^{×} so that 𝔟 = c𝔯. Then for each v ∈ h we can find η_{v} ∈ GL_{2}(K_{v}) such that yc_{v}^{−1}xc_{v} = yxη_{v} for every x ∈ B_{v}. Then y𝔒′_{v} = y𝔒_{v}η_{v}, and so y𝔒′ = Mγ^{−1}η with η = (η_{v})_{v∈h}. Using the map j in the proof of lemma 1.3, we find that η_{v} = diag[1, c_{v}^{ρ}/c_{v}], and so det(γ^{−1}η)𝔯 = φ[y]𝔟^{−1}𝔟^{ρ} ∈ ℑ_{K/F}^{0} . Thus det(ξγ^{−1}η)𝔯 ∈ 𝔞ℑ_{K/F}^{0} , which proves the second assertion. We can put 𝔒 = 𝔯z + 𝔞w with elements z and w. Applying ι to this, we obtain the last assertion.

We call the coset 𝔞ℑ_{K/F}^{0} in the above lemma the characteristic coset of K relative to B. Using this notion, we now reformulate Theorems 2.2 and 2.3 in terms of the matrices Φ in ℌ_{n}^{0}(ξ).

Theorem 2.10 (The case of even n). Let the symbols n, {σ_{v}}_{v∈r_{0}}, ε, and ξ be given as follows: 0 < n ∈ 2Z, σ_{v} ∈ 2Z, |σ_{v}| ≤ n; ε ∈ F^{×}, ξ ∈ 𝔊_{A}. Let B = {K, ε}; let 𝔢 be the product of the prime ideals in F ramified in B, and 𝔨 the characteristic coset of K relative to B. Suppose that (−1)^{σ_{v}/2}ε > 0 at each v ∈ r_{0} and

(2.12) det(ξ)𝔡^{(n−2)/2} ∈ 𝔨.

Then there exists an element Φ of ℌ_{n}^{0}(ξ) such that

(2.13) ε ∈ d_{0}(Φ), det(Φξξ^{ρ})𝔡_{0}^{n/2} = 𝔢, s_{v}(Φ) = σ_{v} for every v ∈ r_{0}.

Moreover, every element of ℌ_{n}^{0}(ξ) is of this type, and the coset T 𝔊ξ and the genus of Φ are determined by (εN_{K/F} (K^{×}), {σ_{v}}_{v∈r_{0}}).

Proof. Let the symbols n, {σ_{v}}_{v∈r_{0}}, ε, and ξ be given as in our theorem. Then Theorem 2.2 combined with Proposition 2.6 guarantees an element Ψ of ℌ_{n}^{0}(η) with some η ∈ 𝔊_{A} such that ε ∈ d_{0}(Ψ) and σ_{v} = s_{v}(Ψ) for every v ∈ r_{0}. Put 𝔶 = det(η)𝔯. By Theorem 2.3 (i), (2.11), and Lemma 2.9 we see that 𝔶𝔡^{(n−2)/2} ∈ 𝔨. Combining this with our condition (2.12), we see that 𝔶 ∈ det(ξ)ℑ_{K/F}^{0} , and so det(η^{−1}ξ) ∈ K^{×}𝔱, which implies η ∈ T 𝔊ξ; see (2.9). By Proposition 2.6 (ii), [Ψ] is isomorphic to [Φ] with some Φ ∈ ℌ_{n}^{0}(ξ). Replacing (Ψ, η) by (Φ, ξ), we obtain (2.13). This proves the first part of our theorem. Conversely, given Φ ∈ ℌ_{n}^{0}(ξ), put L = L_{0}ξ. Let 𝔢 be the product of the prime ideals in F ramified in {K, d_{0}(φ)}. By Theorem 2.2 (i) and (2.11), det(Φξξ^{ρ})𝔡_{0}^{n/2} = 𝔢, which together with Lemma 2.9 implies condition (2.12). This proves the second part. The last part follows from Proposition 2.6.

Theorem 2.11 (The case of odd n). Let the symbols n, {σ_{v}}_{v∈r}, ε, and ξ be given as follows: 0 < n − 1 ∈ 2Z, σ_{v} − 1 ∈ 2Z, |σ_{v}| ≤ n; ε ∈ F^{×} and ξ ∈ 𝔊_{A}. Let ε𝔤 = 𝔞N_{K/F} (𝔟) with an 𝔯-ideal 𝔟 and a squarefree integral 𝔤ideal 𝔞 whose prime factors remain prime in K. Suppose (−1)^{(σ_{v}−1)/2}ε > 0 at each v ∈ r_{0} and

(2.14) det(ξ)𝔡^{(n−1)/2}𝔟 ∈ ℑ_{K/F}^{0} .

Then there exists an element Φ of ℌ_{n}^{0}(ξ) such that

(2.15) ε ∈ d_{0}(Φ), det(Φξξ^{ρ})𝔡_{0}^{(n−1)/2} = 𝔞, s_{v}(Φ) = σ_{v} for every v ∈ r_{0}.

Moreover, every element of ℌ_{n}^{0}(ξ) is of this type, and the coset T 𝔊ξ and the genus of Φ are determined by (εN_{K/F} (K^{×}), {σ_{v}}_{v∈r_{0}}).

This can be proved in exactly the same fashion as for Theorem 2.10.

Lemma 2.12. Suppose F has class number 1. Then the class number of K is odd if and only if ℑ_{K} = ℑ_{K/F}^{0} , in which case −1 ∈ N_{K/Q}(K^{×}) if and only if −1 ∈ N_{K/Q}(𝔯^{×}).

Proof. Suppose the class number of K is odd. Then every 𝔯-ideal 𝔞 is of the form 𝔞 = c𝔟^{2} with an 𝔯-ideal 𝔟 and c ∈ K^{×}. Thus 𝔞 = c𝔟𝔟^{ρ}𝔟(𝔟^{ρ})^{−1} ∈ ℑ_{K/F}^{0} as 𝔟𝔟^{ρ} is principal, and so ℑ_{K} = ℑ_{K/F}^{0} . Suppose the class number of K is even. Then there exists an 𝔯-ideal 𝔵 whose ideal class is not a square. Suppose 𝔵 ∈ ℑ_{K/F}^{0} . Then 𝔵 = z𝔶^{−1}𝔶^{ρ} with z ∈ K^{×} and an 𝔯-ideal 𝔶. Thus 𝔵 = z𝔶𝔶^{ρ}𝔶^{−2}, a contradiction, as 𝔶𝔶^{ρ} is principal. This proves the first part. To prove the second part, suppose −1 = αα^{ρ} with α ∈ K^{×}; put α𝔯 = 𝔟𝔠^{−1} with integral 𝔯deals 𝔟 and 𝔠 that are relatively prime. Then 𝔟𝔟^{ρ} = 𝔠𝔠^{ρ} and we easily see that 𝔟 = 𝔠^{ρ}, and so 𝔠^{2} = α^{−1}𝔠𝔠^{ρ}, which is principal. If the class number of K is odd, then 𝔠 = c𝔯 with c ∈ 𝔯. Thus αc = εc^{ρ} with ε ∈ 𝔯^{×}. Then εε^{ρ} = αα^{ρ} = −1. This completes the proof.

The last statement of the above lemma is false if the class number of K is even. For example, let K = Q(\sqrt{34} ). Then the class number is 2 and −1 = αα^{ρ} with α = (3 + \sqrt{34} )/5, but −1 ∉ N_{K/Q}(𝔯^{×}).

2.13. Let us now take K to be a real or an imaginary quadratic field whose class number is odd. We denote by d the discriminant of K. Thus F = Q and K = Q(\sqrt{d}). By Lemma 2.12 we have ℑ_{K} = ℑ_{K/F}^{0} , and so 𝔊_{A} = T 𝔊 by (2.9). Therefore by Proposition 2.6, every hermitian space over K is isomorphic to [Φ] with Φ ∈ ℌ_{n}^{0}(1_{n}), and 𝔯_{n}^{1} is a maximal lattice in it. For simplicity we put ℌ_{n}^{1} = ℌ_{n}(1_{n}) and ℌ_{n}^{0} = ℌ_{n}^{0}(1_{n}). Then ℌ_{n}^{1} consists of all Φ = (c_{ij}) ∈ ℌ_{n} such that \sqrt{d} c_{ij} ∈ 𝔯 and c_{ii} ∈ Z for every i and j; ℌ_{n}^{0} consists of all Φ ∈ ℌ_{n}^{1} satisfying the following condition:

(2.16) If Φ = P ΨP^{∗}, Ψ ∈ ℌ_{n}^{1}, and P ∈ GL_{n}(K) ∩ M_{n}(𝔯), then det(P) ∈ 𝔯^{×}.

For Φ ∈ ℌ_{n} we put s(Φ) = p − q if K is imaginary and Φ as a complex hermitian matrix has p positive and q negative eigenvalues; we put s(Φ) = 0 if d > 0 and n ∈ 2Z; we do not define s(Φ) if d > 0 and n ∉ 2Z, and so the symbol s(Φ) in that case must be ignored. Clearly two elements Φ_{1} and Φ_{2} of ℌ_{n}^{1} belong to the same genus if s(Φ_{1}) = s(Φ_{2}) and Φ_{1} = P_{v}Φ_{2}P_{v}^{∗} with some P_{v} ∈ GL_{n}(𝔯_{v}) for every v ∈ h. Now, for L = 𝔯_{n}^{1} we have [L̂/L] = det(\sqrt{d} Φ)𝔯 by (2.11). For d > 0 we fix an embedding of K into R, and take \sqrt{d} > 0.

Theorem 2.14 (The case of even n). Let K = Q(\sqrt{d}) as in §2.13, and let three integers n, σ, and e be given as follows: 0 < n ∈ 2Z, σ ∈ 2Z, |σ| ≤ n; σ = 0 if d > 0; e is positive and squarefree. Let r be the number of prime factors of e. Suppose that σ − 2r ∈ 4Z and no prime factor of e splits in K. Then there exists an element Φ of ℌ_{n}^{0} such that

(2.17a) det(\sqrt{d} Φ) = (−1)^{σ/2}e and s(Φ) = σ if d < 0, (2.17b) det(\sqrt{d} Φ) = τe with τ = 1 or −1 if d > 0.

Moreover, every element of ℌ_{n}^{0} is of this type. Its genus is determined by (σ, e) if d < 0, and by e if d > 0. If d > 0 and −1 ∈ N_{K/Q}(K^{×}), then both e and −e can occur as det(\sqrt{d} Φ) for Φ in the same genus. If d > 0 and −1 ∉ N_{K/Q}(K^{×}), then τ is uniquely determined by the condition that a prime number p divides e if and only if τe ∉ N_{K/Q}(K_{p}^{×}).

Proof. Given (n, σ, e) as in our theorem, we can find a quaternion algebra B over Q which is ramified at p if and only if p|e. Then B is definite if and only if r is odd. Since σ − 2r ∈ 4Z, we see that d < 0 if r is odd. Our assumption on the prime factors of e allows us to put B = {K, ε} with ε ∈ Q^{×}. Then (−1)^{σ/2}ε > 0 if d < 0. By Theorem 2.10, we obtain Φ ∈ ℌ_{n}^{0} satisfying (2.13) with ξ = 1_{n}, as (2.12) can be ignored. Then 𝔢 = eZ and det(\sqrt{d} Φ) = τe with τ = ±1. Since s(Φ) = σ, we see that τ = (−1)^{σ/2} if d < 0. The same theorem says that every element of ℌ_{n}^{0} is of this type, and its genus is determined by εN_{K/F} (K^{×}) and σ. We easily see that (e, σ) determines (εN_{K/F} (K^{×}), σ), and vice versa. If −1 ∈ N_{K/Q}(K^{×}), then by Lemma 2.12, 𝔯^{×} contains an element ζ such that ζζ^{ρ} = −1. Then det(P ΦP^{∗}) = − det(Φ) for P = diag[ζ, 1_{n−1}]. Thus both e and −e can happen. Suppose d > 0 and −1 ∉ N_{K/Q}(K^{×}). Then {K, e} is not isomorphic to {K, −e}. Since −d ∈ N_{K/Q}(K^{×}), d_{0}(φ) is represented by det(\sqrt{d}Φ). If det(\sqrt{d}Φ) = τe, then B = \{K, τe\}. Thus τ is uniquely determined by the condition that a prime number p divides e if and only if τe ∉ N_{K/Q}(K_{p}^{×}).

Theorem 2.15 (The case of odd n). Let K = Q(\sqrt{d}) as in §2.13 and let four integers n, σ, τ, and e be given as follows: 0 < n−1 ∈ 2Z; σ is necessary only if d < 0, σ − 1 ∈ 2Z, and |σ| ≤ n; τ is necessary only if d > 0, and τ = 1 or −1; e is positive and squarefree, and every prime factor of e remains prime in K. Then there exists an element Φ of ℌ_{n}^{0} such that

(2.18a) det(\sqrt{d} Φ) = (−1)^{(σ−1)/2}e\sqrt{d} and s(Φ) = σ if d < 0, (2.18b) det(\sqrt{d} Φ) = τe\sqrt{d} if d > 0.

Moreover, every element of ℌ_{n}^{0} is of this type, and its genus is determined by (σ, e) if d < 0 and by (τ, e) if d > 0. For d > 0 the sets (1, e) and (−1, e) determine the same genus if and only if −1 ∈ N_{K/Q}(K^{×}).

Proof. Given (n, σ, τ, e) as in our theorem, take ε = (−1)^{(σ−1)/2}e if d < 0 and ε = (−1)^{(n−1)/2}τe if d > 0. Then Theorem 2.11 with ξ = 1_{n} and 𝔞 = eZ guarantees an element Φ of ℌ_{n}^{0} satisfying (2.15). We can easily verify that (2.18a, b) hold. That every Φ ∈ ℌ_{n}^{0} is of this type also follows from Theorem 2.11, as d_{0}(Φ) can be represented by e or −e with a positive integer e as in our theorem. Since the last assertion is obvious, our proof is complete.

Corollary 2.16. Let K = Q(\sqrt{d}) and ℌ_{n}^{1} be as in §2.13; let 0 < n ∈ Z and σ ∈ Z.

(i) If d < 0, there exists an element Φ of ℌ_{n}^{1} such that det(\sqrt{d} Φ) = 1 and s(Φ) = σ exactly when n ∈ 2Z and σ ∈ 4Z.

(ii) Suppose d > 0; then there exists an element Φ of ℌ_{n}^{1} such that det(\sqrt{d} Φ) = 1 if and only if n ∈ 2Z. Moreover, there exists an element Φ′ of ℌ_{n}^{1} such that det(\sqrt{d} Φ′) = −1 if and only if n ∈ 2Z and −1 ∈ N_{K/Q}(K^{×}).

Proof. From Theorem 2.15 we see that det(\sqrt{d} Φ) = ±1 for Φ ∈ ℌ_{n}^{1} cannot happen if n is odd. Take e = 1 in Theorem 2.14. Then r = 0, and we obtain our results immediately from that theorem. Notice that if d > 0 and −1 ∉ N_{K/Q}(K^{×}), then {K, −1} is a division algebra, and so −1 ∉ N_{K/Q}(K_{p}^{×}) for some prime number p.

The above corollary is a natural analogue of a well-known fact on unimodular quadratic form over Q.

2.17. Examples. (2) Take n = 2, d = 21, and e_{1} = 11 · 13; then the class

number of K is 1 and −1 ∉ N_{K/Q}(K^{×}). For Φ_{1} = [ \begin{matrix}7 & 2/\sqrt{21} \\ −2/\sqrt{21} & −1 \end{matrix} ] we

have det(\sqrt{21} Φ_{1}) = −e_{1} and Φ_{1} ∈ ℌ_{2}^{0}. But we cannot have det(\sqrt{21} Φ) = e_{1} for Φ ∈ ℌ_{2}^{0}.

Next take e_{2} = 3 · 7 · 11 · 13. Then det(\sqrt{21} Φ) for Φ ∈ ℌ_{2}^{0} can be −e_{2} but cannot be e_{2}. Also, {K, 11 · 13} is ramified at p = 3, 7, but {K, −11 · 13} is not. From this we can derive that diag[11, −13] is reduced, but diag[11, 13] is not.

3. Hermitian Diophantine equations over a local field

3.1. Throughout this section we fix (V, φ) in the local case, and put n = dim(V). We denote by 𝔭 the maximal ideal of 𝔤, and by t the core dimension of (V, φ). Then t ≤ 2 as observed in §2.1. For an 𝔯-lattice L in V we put

(3.1) C(L) = {α ∈ U^{φ}(V) | Lα = L}, C^{1}(L) = C(L) ∩ SU^{φ}(V).

Define L[q, 𝔟] and L[q] by (1.10b, c). Clearly L[q, 𝔟] and L[q] are stable under right multiplication by the elements of C(L), and so the four orbit sets L[q, 𝔟]/C(L), L[q]/C(L), L[q, 𝔟]/C^{1}(L), and L[q]/C^{1}(L) are meaningful. Now our principal results of this section are the following two theorems.

Theorem 3.2. Suppose that F is local and n > 1. Let L be a maximal 𝔯-lattice in V. Then for every q ∈ F^{×} and every 𝔯-ideal 𝔟 the following assertions hold:

(i) #{L[q, 𝔟]/C(L)} ≤ 1.

(ii) #{L[q, 𝔟]/C^{1}(L)} < ∞.

(iii) #{L[q, 𝔟]/C^{1}(L)} ≤ 1 if we exclude the following two cases: (a) n = 2 and t = 0; (b) t = 1, q𝔯 = 𝔟𝔟^{ρ}, and 𝔡 ≠ 𝔯.

(iv) If n = 2 and t = 0, then

(3.2)  #\{L[q, 𝔡^{−1}]/C^{1}(L)\} = \{ \begin{matrix}1  if q ∈ 𝔤^{×}, \\ N(q𝔤)[1 − \{K/F \}N(𝔭)^{−1}] if q ∈ 𝔭, \end{matrix}

where N(𝔞) = #(𝔤/𝔞), and {K/F } = 1, −1, or 0, according as K = F × F, K is an unramified quadratic extension of F, or K is ramified over F.

(v) L[q, 𝔡^{−1}] ≠ ∅ exactly in the following cases: (a) t = 0 and q ∈ 𝔤; (b) t = 1, 𝔡 = 𝔯, d_{0}(φ) = N_{K/F} (K^{×}), and q ∈ 𝔤; (c) t = 1, d_{0}(φ) ⊄ 𝔤^{×}N_{K/F} (K^{×}), and q ∈ 𝔭^{−1}; (d) t = 1, 𝔡 ≠ 𝔯, and q ∈ 𝔭𝔡^{−2} ∪ (𝔡^{−2} ∩ d_{0}(φ)); (e) n = t = 2 and 𝔤 ⊂ q𝔤 ⊂ 𝔭^{−1} if 𝔡 = 𝔯; q𝔤 = 𝔭^{−1} if 𝔡 ≠ 𝔯; (f) n > 2 = t and q ∈ 𝔭^{−1}.

The proof will be given in §3.6 through §3.12.

The quantity #{L[q, 𝔟]/C^{1}(L)} in Case (b) of (iii) is not so simple. We will discuss that case in Lemma 3.13 (ii).

For every α ∈ U^{φ} we have C(Lα) = α^{−1}C(L)α, C^{1}(Lα) = α^{−1}C^{1}(L)α, and (Lα)[q, 𝔟] = L[q, 𝔟]α. Therefore, in view of Lemma 1.10 (i), it is sufficient to prove Theorem 3.2 for a special choice of L. Also, since c · L[q, 𝔟] = L[cc^{ρ}q, c𝔟] for every c ∈ K^{×}, it is sufficient to prove our theorem when 𝔟 = 𝔡^{−1} or 𝔟 = 𝔯.

Theorem 3.3. If n > 1, we have #{Λ[q]/C^{1}(Λ)} < ∞ for every q ∈ F^{×} and every 𝔯-lattice Λ in V.

Proof. Let L be a maximal lattice in V. Given Λ, we can find c ∈ F^{×} such that cΛ ⊂ L. Then cΛ[q] ⊂ L[c^{2}q]. For any two open compact subgroups D and E of SU^{φ} we have [D : D ∩ E] < ∞. Therefore it is sufficient to prove that #{L[q]/C^{1}(L)} < ∞. Given h ∈ L[q], put 𝔟 = φ(h, L). Then q𝔯 ⊂ 𝔟 ⊂ 𝔡^{−1}, and hence L[q] ⊂ ⋃_{𝔟} L[q, 𝔟], where 𝔟 runs over the 𝔯-ideals 𝔟 such that q𝔯 ⊂ 𝔟 ⊂ 𝔡^{−1}. Therefore the desired fact follows from Theorem 3.2 (ii).

Lemma 3.4. Let L be a maximal lattice in V. Suppose dim(V) > 1 and q ∈ 𝔤^{×}; then #{L[q, 𝔯]/C(L)} ≤ 1. Moreover, #{L[q, 𝔯]/C^{1}(L)} ≤ 1 if K is a field unramified over F, or the core dimension of (Kh)^{⊥} is not 0 for some h ∈ L[q, 𝔯].

Proof. Let h, k ∈ L[q, 𝔯]. We see that L + 𝔯h is integral, and so h ∈ L, as L is maximal. Given x ∈ L, put y = x − φ[h]^{−1}φ(x, h)h. Then y ∈ L ∩ (Kh)^{⊥}. From this we can derive that L = 𝔯h ⊕ M with M = L ∩ (Kh)^{⊥}. Similarly L = 𝔯k ⊕ M′ with M′ = L ∩ (Kk)^{⊥}. Since L is maximal, M and M′ must be maximal. By (1.5) we can find an element α of SU^{φ}(V) such that k = hα. Then Mα is a maximal lattice in (Kk)^{⊥}. By Lemma 1.10 (i) we can find an element β of U^{φ}((Kk)^{⊥}) such that Mαβ = M′; by Lemma 1.10 (ii) such a β can be taken from SU^{φ}((Kk)^{⊥}) if K is a field unramified over F, or the core dimension of (Kk)^{⊥} is not 0. Extend β to V by putting kβ = k. Then αβ ∈ C(L) and hαβ = k. We have αβ ∈ C^{1}(L) under the said conditions on K or on Kk. This proves our lemma. Notice that the core dimension of (Kh)^{⊥} depends only on hU(φ).

3.5. We call an element x of 𝔯_{n}^{1} primitive if x𝔯_{1}^{n} = 𝔯. Replacing 𝔯 by 𝔤, we can similarly define the primitive elements of 𝔤_{n}^{1}. Given an integral 𝔯-lattice L in V, identify V and L with K_{n}^{1} and 𝔯_{n}^{1} with respect to an 𝔯-basis {z_{i}}_{i=1}^{n} of L; also let φ_{0} = (φ(z_{i}, z_{j}))_{i,j=1}^{n}. By (1.9) we have δφ_{0} ≺ 𝔯 for any element δ of 𝔯 such that δ𝔯 = 𝔡. Moreover, φ(x, L) = 𝔡^{−1} for x ∈ L = 𝔯_{n}^{1} if and only if δxφ_{0} is primitive.

To prove Theorem 3.2, we fix a maximal lattice L in V, and hereafter write simply C and C^{1} for C(L) and C^{1}(L); we always assume n > 1.

3.6. In this subsection we consider the case K = F × F, using the notation of §1.8. We can put L = 𝔤_{n}^{1} × 𝔤_{n}^{1}.

Let h = (a, b) ∈ L[q, 𝔯] with a, b ∈ F_{n}^{1}. Then φ(h, L) = (a𝔤_{1}^{n}, b𝔤_{1}^{n}), and so both a and b are primitive. Let {e_{i}}_{i=1}^{n} be the standard basis of F_{n}^{1}. Take α ∈ SL_{n}(𝔤) so that aα = e_{1} and put c = b · ^{t}α^{−1} = (c_{i})_{i=1}^{n}. Then

c_{1} = q. Suppose q ∈ 𝔤^{×}; then define β ∈ SL_{n}(𝔤) by β = [ \begin{matrix}1 & x \\ 0 & 1_{n−1} \end{matrix} ] with

x = (q^{−1}c_{2}, …, q^{−1}c_{n}) and put γ = α · ^{t}β. Then aγ = e_{1} and b · ^{t}γ^{−1} = cβ^{−1} = qe_{1}. Thus L[q, 𝔯]/C^{1} is represented by (e_{1}, qe_{1}) if q ∈ 𝔤^{×}.

Next suppose q ∈ 𝔭; then (c_{2}, …, c_{n}) is primitive, and so we can find β ∈ GL_{n−1}(𝔤) such that (c_{2}, …, c_{n})β = (1, 0, …, 0). Put γ = α · diag[1, ^{t}β^{−1}]. Then aγ = e_{1} and b·^{t}γ^{−1} = (q, 1, 0, …, 0). This shows that #{L[q, 𝔯]/C} = 1. If n > 2, we can take β ∈ SL_{n−1}(𝔤), and hence #{L[q, 𝔯]/C^{1}} = 1.

Finally suppose n = 2 and q ∈ 𝔭; then we have shown that L[q, 𝔯]/C^{1} can be represented by the elements of the form (e_{1}, (q, s)) with s ∈ 𝔤^{×}. Suppose (e_{1}, (q, s))α = (e_{1}, (q, t)) with α = (γ, ^{t}γ^{−1}) ∈ C^{1}, γ ∈ SL_{2}(𝔤), and s, t ∈

𝔤^{×}. Clearly γ = [ \begin{matrix}1 & 0 \\ v & 1 \end{matrix} ] with v ∈ 𝔤, and so t = s − qv. Since the procedure

is reversible, we see that #{L[q, 𝔯]/C^{1}} = #(𝔤/q𝔤)^{×}, which gives (3.2) for K = F × F. This completes the proof of Theorem 3.2 in the case K = F × F.

Hereafter from §3.7 through §3.12 we assume that K is a field.

3.7. Let us consider the case n = t = 2. Let the symbols be as in Lemma 1.3 and §1.11; we identify B with j(B). In view of Lemma 1.6 we can take φ[ℓ] = c = 1 in the proof of Lemma 1.3, and so we can take γ = 1 in §1.11. Thus M = ℓ𝔒. Since φ is anisotropic, SU^{φ}(V) = C^{1}(M) = {α ∈ 𝔒^{×} | αα^{ι} = 1}. From (1.5) we see that #{M[q, 𝔡^{−1}]/C^{1}} = 1 if M[q, 𝔡^{−1}] ≠ ∅. Since Tr_{K/F} (φ(ℓα, ℓβ)) = Tr_{B/F} (αβ^{ι}) for α, β ∈ B, we have φ(ℓα, M) = 𝔡^{−1} only if Tr_{B/F} (α𝔒) = 𝔤, which is so if and only if α𝔒 = 𝔒 or α𝔒 = 𝔓^{−1}. Take such an α and assume that K is unramified over F. Then ℓα ∈ M[αα^{ι}, 𝔯]. Thus M[q, 𝔡^{−1}] ≠ ∅ if and only if q𝔤 is 𝔤 or 𝔭^{−1}. Next suppose that K is ramified over F. Clearly M[q, 𝔡^{−1}] ≠ ∅ for some q, and q𝔤 is 𝔤 or 𝔭^{−1} for the same reason as above. Suppose q ∈ 𝔤^{×}. Then we can find an element ξ ∈ 𝔒^{×} such that ξξ^{ι} = q. Then φ[xξ] = qφ[x] for every x ∈ V, Mξ = M, and M[1, 𝔟]ξ = M[q, 𝔟]. Let d_{0}(φ) = sN_{K/F} (K^{×}) as in the proof of Lemma 1.3. We may assume that s ∈ 𝔤^{×}, s ∉ N_{K/F} (𝔯^{×}), and φ_{0} = diag[1, −s]. Observe that 𝔒 consists of the elements a + bω with a, b ∈ K such that a + a^{ρ} ∈ 𝔤 and aa^{ρ} − sbb^{ρ} ∈ 𝔤. Now let ℓα ∈ M[1, 𝔡^{−1}] with α ∈ B. Then αα^{ι} = 1, and so ℓ ∈ M[1, 𝔡^{−1}]. Thus 𝔡^{−1} = φ(ℓ, M) = φ(ℓ, ℓ𝔒). For a + bω ∈ 𝔒 as above, we have φ(ℓ, ℓ(a + bω)) = a^{ρ}, and so 𝔒 contains an element a + bω such that a𝔯 = 𝔡^{−1}. Put N_{K/F} (𝔡) = 𝔭^{κ} with 0 < κ ∈ Z. Then aa^{ρ}𝔤 = 𝔭^{−κ}, and bb^{ρ}𝔤 = 𝔭^{−κ} as aa^{ρ} −sbb^{ρ} ∈ 𝔤. Putting a^{−1}b = c, we obtain scc^{ρ} −1 ∈ 𝔭^{κ}, a contradiction, as s ∉ N_{K/F} (𝔯^{×}). Thus M[q, 𝔡^{−1}] ≠ ∅ only if q𝔤 = 𝔭^{−1}. This proves Theorem 3.2 when n = t = 2.

3.8. Take an element u ∈ 𝔯 so that 𝔯 = 𝔤[u] and put δ = u − u^{ρ}. Then 𝔡 = δ𝔯 and δ^{ρ} = −δ. We will often use these u and δ in our later treatment.

Take a decomposition of V as in (1.8a, b), and assume that it is a Witt decomposition; thus t = dim(Z). Put g_{i} = δ^{−1}f_{i} and

(3.3)  L = ∑_{i=1}^{r}(𝔯e_{i} + 𝔯g_{i}) + M,  M = {y ∈ Z | φ[y] ∈ 𝔤}.

Then L is a maximal lattice in V as noted in §1.7. With an 𝔯basis {m_{i}}_{i=1}^{t} of M, consider matrix representation with respect to {e_{1}, …, e_{r}, m_{1}, …, m_{t}, g_{1}, …, g_{r}}. Then φ is represented by

(3.4) φ_{0} = [ \begin{matrix}0 & 0 & −δ^{−1}1_{r} \\ 0 & ζ & 0 \\ δ^{−1}1_{r} & 0 & 0 \end{matrix} ] ,

where ζ = (φ(m_{i}, m_{j}))_{i,j=1}^{t}. Now write an element of GL(V) as a matrix with 9 matrix blocks corresponding to the blocks of (3.4), and let P be the group consisting of the elements of U^{φ} whose lower left 3 blocks under the diagonal blocks are all 0; also let P^{1} = P ∩ SU^{φ}. Then

(3.5) U^{φ} = P C(L) and SU^{φ} = P^{1}C^{1}(L);

see [S2, Proposition 5.16]. If r = 1, P consists of the matrices of the form

(3.6)  [ \begin{matrix}a & b & a^{−ρ}(s + u^{ρ}bζb^{∗}) \\ 0 & e & −δa^{−ρ}eζb^{∗} \\ 0 & 0 & a^{−ρ} \end{matrix} ] ,

where a ∈ K^{×}, b ∈ K_{t}^{1}, e ∈ U^{ζ}(Z), and s ∈ F.

3.9. The case t = 0. Represent the elements of V by row vectors in K_{r}^{1} ×K_{r}^{1} with respect to the basis {e_{1}, …, e_{r}, g_{1}, …, g_{r}}, and GL(V) by GL_{2r}(K), acting on the right. Then for h = (y, z) ∈ K_{r}^{1} × K_{r}^{1} we have φ[h] = δ^{−1}(zy^{∗} − yz^{∗}) and φ(h, L) = δ^{−1} ∑_{i=1}^{r}(𝔯y_{i}+𝔯z_{i}). Now φ is represented by δ^{−1} η, where

η = [ \begin{matrix}0 & −1_{r} \\ 1_{r} & 0 \end{matrix} ] .

Therefore diag[α, α̃] ∈ U^{φ} for every α ∈ GL_{r}(K), where α̃ = (α^{∗})^{−1}. Suppose φ(h, L) = 𝔡^{−1} for h = (y, z); then ∑_{i=1}^{r}(𝔯y_{i} + 𝔯z_{i}) = 𝔯 and φ[h] ∈ 𝔤. Putting k = (e_{1}, qug_{1}) with q ∈ F (not necessarily ≠ 0), let us prove

(3.7) φ[h] = q and φ(h, L) = 𝔡^{−1} ⟹ h ∈ kC(L); h ∈ kC^{1}(L) if r > 1.

Since η ∈ C^{1}, changing h for hη if necessary, we may assume that ∑_{i=1}^{r} 𝔯y_{i} = 𝔯. We can find an element α ∈ GL_{r}(𝔯) such that yα = e_{1}; we can even take α from SL_{r}(𝔯) if r > 1. Put w = zα̃. Then (y, z)diag[α, α̃] = (e_{1}, w), and so w_{1} − w_{1}^{ρ} = qδ. Thus we can put w_{1} = p + qu with p ∈ 𝔤. Define an element s = s^{∗} ∈ M_{r}(𝔯) so that s_{11} = p and s_{1j} = w_{j} for j > 1, and put β = [ \begin{matrix}1_{r} & s \\ 0 & 1_{r} \end{matrix} ] . Then β ∈ C^{1} and kβ = (e_{1}, w). If r > 1, then we see that

h ∈ kC^{1}, and so L[q, 𝔡^{−1}] = kC^{1}. This proves (3.7) and also Theorem 3.2 (i), (iii), (v) when t = 0.

Suppose r = 1; then y ∈ 𝔯^{×}. Since C^{1} is a normal subgroup of C, the above argument shows that h ∈ ⋃_{t}(t, t^{−ρ}qu)C^{1}, where t runs over 𝔯^{×}. Define B as in Lemma 1.3. From (1.7b), (1.15), and the last equality in the proof of Lemma 1.3 we obtain B = M_{2}(F), SU^{φ} = SL_{2}(F), and C^{1} = SL_{2}(𝔤). Let t ∈ 𝔯^{×} ∩ 𝔤[qu]. Then t = a + cqu with a, c ∈ 𝔤, and we can find an element

γ ∈ SL_{2}(𝔤) of the form γ = [ \begin{matrix}a & b \\ c & d \end{matrix} ] . We have (1, qu)γ = (t, t^{−ρ}v) with

v ∈ 𝔯. Then v − v^{ρ} = qδ, so that v = qu − s with s ∈ 𝔤. Since tt^{ρ} ∈ 𝔤^{×},

we can put s = tt^{ρ}s′ with s′ ∈ 𝔤. Put σ = [ \begin{matrix}1 & s′ \\ 0 & 1 \end{matrix} ] . Then σ ∈ C^{1} and

(t, t^{−ρ}v)σ = (t, t^{−ρ}qu), which shows that (t, t^{−ρ}qu) ∈ (1, qu)C^{1} if t ∈ 𝔤[qu]. Since the converse is obvious, we thus obtain L[q, 𝔡^{−1}] = ⨆_{t∈τ} (t, t^{−ρ}qu)C^{1}, where τ = 𝔯^{×}/𝔤[qu]^{×}. This proves (3.2) and completes the proof of Theorem 3.2 when t = 0.

3.10. The notation being as in (3.3), put H_{j} = ∑_{i=1}^{j}(Ke_{i} + Kg_{i}). Let us now show that given h ∈ V, there exists an element α of C^{1} such that hα ∈ H_{1} + Z. This is obvious if h ∈ Z or r = 1. So assume that h ∉ Z and r > 1. Put h = w +k with k ∈ Z and w = ∑_{i=1}^{r}(y_{i}e_{i} +z_{i}g_{i}) ∈ H_{r}, y_{i}, z_{i} ∈ K. Then we can put ∑_{i=1}^{r}(y_{i}𝔯 + z_{i}𝔯) = d𝔯 with d ∈ K^{×}. Taking d^{−1}w as h of (3.7), we can find an element γ ∈ C^{1}(L ∩ H_{r}) such that d^{−1}wγ ∈ H_{1}. Extend γ to an element of C^{1}(L) by defining xγ = x for every x ∈ Z. Then we obtain the desired fact. This means that if n > t > 0, then it is sufficient to prove Theorem 3.2 when r = 1.

3.11. Case t > 0, r = 1. Writing simply e and g for e_{1} and g_{1}, we have V = Ke + Z + Kg and L = 𝔯e + M + 𝔯g with a maximal lattice M in Z. Let h = ye + x + zg with y, z ∈ K and x ∈ Z. Then φ[h] = δ^{−1}(zy^{ρ} − yz^{ρ}) + ζ[x], where ζ is the restriction of φ to Z, and φ(h, L) = δ^{−1}(𝔯y + 𝔯z) + ζ(x, M). Suppose h ∈ L[q, 𝔡^{−1}]. Then 𝔯y+𝔯z+𝔡ζ(x, M) = 𝔯, and hence y, z ∈ 𝔯, x ∈ M̂, and φ[h] − ζ[x] ∈ 𝔤. We identify an element of Z with a row vector of K_{t}^{1} with respect to an 𝔯-basis of M. Then an element ae+b+cg of V with a, c ∈ K and b ∈ Z can be identified with a row vector [a b c] of K_{t+2}^{1}. If t = 2 and M corresponds to 𝔒 as in §1.11, then M̂ corresponds to 𝔓^{−1}, and so ζ[x] ∈ 𝔭^{−1}; consequently φ[h] ∈ 𝔭^{−1}. Since e + qug ∈ L[q, 𝔡^{−1}] if q ∈ 𝔤, we see that L[q, 𝔡^{−1}] ≠ ∅ for every q ∈ 𝔤.

(a) First suppose t = 2 and q ∈ 𝔭. Suppose also that 𝔯y + 𝔯z ≠ 𝔯. Then yz^{ρ} ∈ 𝔭𝔯, and so δ^{−1}(zy^{ρ} −yz^{ρ}) ∈ 𝔭. Thus ζ[x] ∈ 𝔭, and hence ζ(x, M) ≠ 𝔡^{−1}. (In §3.7 we showed that ζ(x, M) = 𝔡^{−1} only if 𝔤 ⊂ φ[x]𝔤 ⊂ 𝔭^{−1}.) Therefore δφ(h, L) ≠ 𝔯, a contradiction. This shows that (y, z) must be primitive. By (3.7) there is an element α of C(𝔯e + 𝔯g) such that (ye + zg)α = e + aug with a ∈ 𝔤. Extend α to an element γ of SU^{φ}(V) by defining wγ = wσ for w ∈ Z with σ ∈ U^{φ}(Z) such that det(σ) = det(α)^{−1}. Since Mσ = M, we see that γ ∈ C^{1}, and hγ = [1 k au] with k ∈ M such that ζ[k] = q − a. Given another h′ ∈ L[q, 𝔡^{−1}], we can similarly find an element γ′ of C^{1} such that h′γ′ = [1 k′ a′u] with a′ ∈ 𝔤 and k′ ∈ M such that ζ[k′] = q − a′. Put b = k′ − k and

(3.8) τ = [ \begin{matrix}1 & b & s + u^{ρ}bζb^{∗} \\ 0 & 1 & −δζb^{∗} \\ 0 & 0 & 1 \end{matrix} ]

with some s ∈ 𝔤. This is a special case of (3.6) and belongs to SU^{φ}(V). Since b ≺ 𝔯 and δζ ≺ 𝔯, we see that τ ∈ C^{1}. We choose s so that hγτ = [1 k′ a′u], which is so if and only if a′u = s + u^{ρ}bζb^{∗} − δkζb^{∗} + au. This can be achieved by taking s = Tr_{K/F} (uζ(k, k′)) − (u + u^{ρ})ζ[k′]. Then h′γ′ = hγτ, and so h′ ∈ hC^{1} as expected.

(b) Next suppose t = 2 and q ∉ 𝔭. Then q ∈ 𝔤^{×} or q𝔤 = 𝔭^{−1}. If 𝔡 ≠ 𝔯 and q ∈ 𝔤, then ζ[x] ∈ 𝔤, and so δζ(x, M) ≠ 𝔯 as shown in §3.7. Consequently 𝔯y+𝔯z = 𝔯 in such a case, and the argument of case (a) is applicable. Therefore we may assume that q𝔤 = 𝔭^{−1} if 𝔡 ≠ 𝔯. Then as observed in §3.7, we can find an element v of M[q, 𝔡^{−1}]. The same can be said for both cases q ∈ 𝔤^{×} and q𝔤 = 𝔭^{−1} if 𝔡 = 𝔯. We identify v with the row vector [0 v 0], which can be viewed as an element of L[q, 𝔡^{−1}], and so L[q, 𝔡^{−1}] ≠ ∅ in such cases. Combining (1.5) and (3.5), we have h ∈ vP^{1}C^{1}, and hence h = vπα with π ∈ P^{1} and α ∈ C^{1}. Write π in the form (3.6) and focus our attention on the element e of U^{ζ}(Z) there. Since U^{ζ}(Z) = C(M), we see that ve ∈ M[q, 𝔡^{−1}], and hence ve = vε with ε ∈ C^{1}(M) = SU^{ζ}(Z) as shown in §3.7. Let β = diag[1, ε, 1] and π_{1} = πβ^{−1}. Then β ∈ C^{1}, π_{1} ∈ P^{1}, and vπ_{1} = hα^{−1}β^{−1} ∈ L[q, 𝔡^{−1}]. Our choice of ε shows that vπ_{1} = [0 v p] with p ∈ 𝔯. Since v ∈ M[q, 𝔡^{−1}], δvζ is primitive (see §3.5), and so we can find an element b of 𝔯_{2}^{1} such that δvζb^{∗} = −p. Define τ by (3.8) with this b and s = 0. Then τ ∈ C^{1} and vτ = vπ_{1} = hα^{−1}β^{−1}. This shows that h ∈ vC^{1}. Thus we obtain Theorem 3.2 when n > t = 2.

(c) Finally suppose t = 1; let M and ζ be as in (3.3) and (3.4). Then ζ = φ[m] and M = 𝔯m with an element m. Clearly ζ ∈ 𝔤^{×} or ζ𝔤 = 𝔭; the latter case occurs only when 𝔡 = 𝔯. We first treat the case where ζ ∈ 𝔤^{×}; the other case will be treated in §3.12. Thus h = ye+zg+sm with y, z, s ∈ K such that y𝔯+z𝔯+s𝔡 = 𝔯 and δ^{−1}(zy^{ρ} −yz^{ρ})+ζss^{ρ} = q. Suppose 𝔡 = 𝔯 and y𝔯+z𝔯 ≠ 𝔯. Then s ∈ 𝔯^{×} and we see that q = φ[h] ∈ 𝔤^{×}. Then #{L[q, 𝔡^{−1}]/C^{1}(L)} ≤ 1 by Lemma 3.4.

(c1) Let us now prove the case in which q ∈ 𝔤 and y𝔯 + z𝔯 = 𝔯 for both ramified and unramified K. Putting p = δ^{−1}(zy^{ρ} − yz^{ρ}) and applying (3.7) to ye + zg, we find γ ∈ C(L) such that Mγ = M and (y, z)γ = (1, pu) with p∈𝔤. Replacing γ by γα with α∈C such that mα=det(γ)^{−1}m and α is the identity map on Ke+Kg, we may assume that γ ∈C^{1}. We have hγ = [1 x pu] with x ∈ 𝔯. We consider

(3.9) σ = [ \begin{matrix}1 & −x & u^{ρ}ζxx^{ρ} \\ 0 & 1 & δζx^{ρ} \\ 0 & 0 & 1 \end{matrix} ] ,

which is similar to (3.8) and belongs to C^{1}. We have then hγσ = [1 0 s] with s ∈ 𝔤. Taking (1, s) as (y, z) above, we find γ′ ∈ C^{1} such that hγσγ′ = [1 0 qu]. Thus h ∈ kC^{1} with k = e + qug; also Theorem 3.2 (v), (b) is valid.

(c2) Suppose K is ramified over F. If q ∈ 𝔤, then s ∈ 𝔯, so that y𝔯+z𝔯 = 𝔯. and (c1) covers this case. Thus we assume that q ∉ 𝔤. Then s ∉ 𝔯 and q𝔤 = ss^{ρ}𝔤. Put 𝔡 = 𝔮^{κ} with the maximal ideal 𝔮 of 𝔯 and 0 < κ ∈ Z. Since s𝔡 ⊂ 𝔯, we can put s^{−1}𝔯 = 𝔮^{a} with 0 < a ≤ κ; then q𝔤 = 𝔭^{−a}. Thus a is determined by q. Suppose a < κ; then s𝔡 ≠ 𝔯, so that y𝔯 + z𝔯 = 𝔯. By the same technique as in (c1), we can find γ ∈ C^{1} such that hγ = [1 x pu] with p ∈ 𝔤 and x ∈ K. Then x𝔮^{a} = 𝔯. Let k = [1 x_{1} p_{1}u] ∈ L[q, 𝔡^{−1}] with p_{1} ∈ 𝔤 and x_{1} such that x_{1}𝔮^{a} = 𝔯. Our task is to show that k ∈ hC. For simplicity put N(w) = ww^{ρ} for w ∈ K^{×}. We have ζN(x) + p = q = ζN(x_{1}) + p_{1}, and so N(x^{−1}x_{1}) − 1 ∈ 𝔭^{a}. Thus N(x^{−1}x_{1}) ∈ N(𝔯^{×}) ∩ (1 + 𝔭^{a}) = N(1 + 𝔮^{a}) by [S2, Lemma 17.6 (2)]. We can therefore put N(x^{−1}x_{1}) = N(d) with d ∈ 1 + 𝔮^{a}. Put α = diag[1_{r}, dx_{1}^{−1}x, 1_{r}]. Then α ∈ C^{1} and kα = [1 dx p_{1}u]. Put b = dx − x and consider τ of (3.8) with this b and any s ∈ 𝔤. Then τ ∈ C^{1} and hγτ = [1 dx c] with c ∈ 𝔤 such that c − c^{ρ} = δp_{1}. Choosing s suitably, we obtain c = p_{1}u. Then k ∈ hC^{1}.

(c3) It remains to treat the case a = κ. Then s𝔡 = 𝔯 and q𝔯 = 𝔡^{−2}. Put h_{0} = δh and q_{0} = −δ^{2}q. Then h_{0} ∈ L[q_{0}, 𝔯]. Since q_{0} ∈ 𝔤^{×}, by Lemma 3.4 we see that #{L[q, 𝔡^{−1}]/C} = #{L[q_{0}, 𝔯]/C} ≤ 1.

(c4) As for (d) of Theorem 3.2 (v), we have seen that L[q, 𝔡^{−1}] ≠ ∅ only if q ∈ 𝔡^{−2}; also e + qug ∈ L[q, 𝔡^{−1}] if q ∈ 𝔤. Thus it remains to consider the case where 𝔡 ≠ 𝔯 and q ∉ 𝔤. Suppose q𝔤 = 𝔭^{−a} with 0 < a < κ. We can find c ∈ 𝔤^{×} such that c − 1 ∈ 𝔭^{a} and c ∉ N(𝔯^{×}). Then q or c^{−1}q represents d_{0}(φ). If q ∈ d_{0}(φ), then q = ζss^{ρ} with s ∈ K, and e + sm ∈ L[q, 𝔡^{−1}]. If q ∈ cd_{0}(φ), put q = cζss^{ρ} with s ∈ K and p = ζss^{ρ}(c − 1). Then p ∈ 𝔤 and e + pu𝔤 + sm ∈ L[q, 𝔡^{−1}]. Thus L[q, 𝔡^{−1}] ≠ ∅ if q ∈ 𝔭𝔡^{−2}. The case q𝔯 = 𝔡^{−2} will be settled in Lemma 3.13.

3.12. Let us now treat the case where t = 1 and ζ𝔤 = 𝔭. Then 𝔡 = 𝔯 and δ ∈ 𝔯^{×}. To avoid possible confusion, we use the letter π instead of ζ; thus φ[m] = π. Let h = ae + bf + cm ∈ L[q, 𝔯]. Then 𝔯a + 𝔯b + 𝔯πc = 𝔯 and ab^{ρ} + a^{ρ}b + πcc^{ρ} = q. Clearly q ∈ 𝔭^{−1}; also c ∈ 𝔯 if and only if q ∈ 𝔤. Given q ∈ 𝔤, we can find b ∈ 𝔯 such that b + b^{ρ} = q. Then e + bf ∈ L[q, 𝔯]. Given q ∈ 𝔭^{−1}, ∉ 𝔤, we can find d ∈ 𝔯^{×} such that dd^{ρ} = πq. Then π^{−1}dm ∈ L[q, 𝔯]. Thus L[q, 𝔯] ≠ ∅ if and only if q ∈ 𝔭^{−1}.

(1) Let us first assume that q ∈ 𝔤; then c ∈ 𝔯 and 𝔯a + 𝔯b = 𝔯. By (3.7) there exists an element β of C(𝔯e + 𝔯f) such that kβ = e + sf with s ∈ 𝔯. Extend β to an element α of C^{1}(L) by defining mα = det(β)^{−1}m. Then hα = e + sf + c_{1}m with c_{1} ∈ 𝔯. Now represent the elements of U^{φ}(V) by matrices with respect to {e, m, f}. Then (3.5) holds with the subgroup P of U^{φ} consisting of the upper triangular matrices. Observe that P contains every matrix of the form

(3.10)  [ \begin{matrix}1 & −x & y \\ 0 & 1 & πx^{ρ} \\ 0 & 0 & 1 \end{matrix} ]

with x, y ∈ K such that y + y^{ρ} = −πxx^{ρ}. Since Tr_{K/F} (𝔯) = 𝔤, for any x ∈ 𝔯 we can take such a y from 𝔯. Let γ be the matrix of (3.10) with x = c_{1} and y ∈ 𝔯. Then γ ∈ C^{1}(L) and hαγ = e + zf with z ∈ 𝔯 such that z + z^{ρ} = q. Take z_{1} ∈ 𝔯 so that z_{1} + z_{1}^{ρ} = q. Denote by ε the matrix of (3.10) with x = 0 and y = z_{1} −z. Then ε ∈ C^{1}(L) and (e+zf)ε = e+z_{1}f. This gives the desired result when q ∈ 𝔤.

(2) Next we consider the case c ∉ 𝔯; put d = πc. Then d ∈ 𝔯^{×} and q = ab^{ρ} + a^{ρ}b + π^{−1}dd^{ρ}. Thus πq ∈ 𝔤^{×}, and so we can find an element d_{0} ∈ 𝔯^{×} such that d_{0}d_{0}^{ρ} = πq. Put k = π^{−1}d_{0}m. Then k ∈ L[q, 𝔯]. By (1.5), we have h = kα with α ∈ SU^{φ}, and by (3.5) we can put α = βγ with β ∈ P^{1} and γ ∈ C^{1}. Replacing β by βξ with a suitable diagonal matrix ξ belonging to C^{1}, we may assume that the center entry of β is 1. (Here we need Lemma 1.9 (i).) Let [0 1 j] be the second row of β. Then [0 π^{−1}d_{0} π^{−1}d_{0}j] = kβ = hγ^{−1} ∈ L[q, 𝔯], and so π^{−1}d_{0}j ∈ 𝔯. Put x = π^{−1}j^{ρ}; then x ∈ 𝔯. Let ε be the matrix of (3.10) with this x and y such that y + y^{ρ} = −πxx^{ρ}. Then ε ∈ C^{1} and kε = kβ = hγ^{−1}, which gives the desired fact.

It only remains to discuss L[q, 𝔡^{−1}]/C^{1} when 𝔡 ≠ 𝔯 and q𝔯 = 𝔡^{−2}. (In (c3) we treated L[q, 𝔡^{−1}]/C.) The problem is settled by (ii) of the following Lemma.

Lemma 3.13. If K is a field ramified over F, then the following assertions hold:

(i) Let W = (Kh)^{⊥} with h ∈ L[q, 𝔟] and let E_{1} = det (C ∩ U^{φ}(W)), where we view U^{φ}(W) as the subgroup of U^{φ}(V) consisting of the elements ξ such that hξ = h. Then #{L[q, 𝔟]/C^{1}} = [E_{L} : E_{1}] < ∞, where E_{L} is as in Lemma 1.9.

(ii) If 1 < n − 1 ∈ 2Z and q𝔯 = 𝔡^{−2}, then #{L[q, 𝔡^{−1}]/C^{1}} = 2 or 0 according as q represents d_{0}(φ) or not.

Proof. Clearly C∩U^{φ}(W) is an open subgroup of U^{φ}(W). Now det : U^{φ}(W) → E is a continuous surjective map, and so it is an open map by virtue of a well know principle; see [S3, Lemma 8.0]. Thus E_{1} is an open subgroup of E, and so [E : E_{1}] < ∞, as E is compact. Therefore [E_{L} : E_{1}] < ∞. Next, take a finite subset B ⊂ C so that {det(β)}_{β∈B} gives E_{L}/E_{1}. Let k ∈ L[q, 𝔟]. By Theorem 3.2 (i), k = hα with α ∈ C. Then det(αβ^{−1}) ∈ E_{1} for some β ∈ B, so that det(αβ^{−1}) = det(γ) with γ ∈ C ∩ U^{φ}(W). Put ξ = α^{−1}γβ. Then ξ ∈ C^{1} and hβ = hγβ = kξ, that is, k ∈ hβC^{1}. Thus L[q, 𝔟] = ⋃_{β∈B} hβC^{1}. We easily see that the last union is disjoint, and so we obtain (i). To prove (ii), let the notation be as in (3.3) and (3.4); let M = 𝔯m, φ[m] = ζ ∈ 𝔤^{×}, and 𝔡 = 𝔮^{κ} as in (c2) above. Then ζ represents d_{0}(φ). Suppose q𝔯 = 𝔡^{−2} and h ∈ L[q, 𝔡^{−1}]; put h = ∑_{i=1}^{r}(y_{i}e_{i} + z_{i}g_{i}) + sm with y_{i}, z_{i}, s ∈ K. Then y_{i}, z_{i} ∈ 𝔯, and q − ζss^{ρ} ∈ 𝔤, so that s𝔯 = 𝔡^{−1} and (ζss^{ρ})^{−1}q ∈ 1 + 𝔭^{κ} ∈ N_{K/F} (𝔯^{×}); thus q represents d_{0}(φ). This shows that L[q, 𝔡^{−1}] ≠ ∅ only for such a q. Taking such a q, we can put q = ζxx^{ρ} with x ∈ K^{×}. Let k = xm and W = (Kk)^{⊥}. Then k ∈ L[q, 𝔡^{−1}], W = ∑_{i=1}^{r}(Ke_{i} + Kg_{i}), and C ∩ U^{φ}(W) = C(Λ) with Λ = ∑_{i=1}^{r}(𝔯e_{i} + 𝔯g_{i}); thus E_{1} = E_{Λ}. By Lemma 1.9, E_{L} = E, E_{Λ} = E_{0}, and [E : E_{0}] = 2, which together with (i) shows that #{L[q, 𝔡^{−1}]/C^{1}} = 2. This completes the proof.

Lemma 3.14. Let W = (Kh)^{⊥} with h ∈ L[q, 𝔟] and let Λ = L∩W. Suppose that 0 < n − 1 ∈ 2Z. Then there is a unique maximal lattice in W containing Λ at least in the following two cases: (1) q(𝔟𝔟^{ρ})^{−1} = 𝔯; (2) q(𝔟𝔟^{ρ})^{−1} = 𝔭𝔯 and {K, qd_{0}(φ)} is a division algebra. Moreover, Λ is maximal and C(Λ) = C(L)∩ U^{φ}(W) in Case (1); Λ is maximal and [C(Λ) : C(L) ∩ U^{φ}(W)] = N(𝔭) + 1 in Case (2) if K is unramified over F and d_{0}(φ) = N_{K/F} (K^{×}). These assertions are true with C^{1} instead of C.

Proof. Changing h for ch with some c ∈ K^{×}, we may assume that 𝔟 = 𝔯. This does not change the ideal q(𝔟𝔟^{ρ})^{−1} nor {K, qd_{0}(φ)}. Thus q ∈ 𝔤^{×} in Case (1) and q𝔤 = 𝔭 in Case (2). Suppose q ∈ 𝔤^{×}. Then L = 𝔯h ⊕ Λ as shown in the proof of Lemma 3.4, and Λ is maximal as noted there; clearly C(Λ) = C(L) ∩ U^{φ}(W) and C^{1}(Λ) = C^{1}(L) ∩ U^{φ}(W). Next suppose q𝔤 = 𝔭 and {K, qd_{0}(φ)} is a division algebra. Let u, δ, and {e_{i}, g_{i}} be as in §3.8. Since n ∉ 2Z, we have L = 𝔯m + ∑_{i=1}^{r}(𝔯e_{i} + 𝔯g_{i}) with an element m such that φ[m] = ζ. Theorem 3.2 (i) allows us to replace h by any element in L[q, 𝔯]. Thus we can put h = e_{1} + qug_{1}. Put now k = e_{1} + qu^{ρ}g_{1} , Y = Kk + Km, and N = 𝔯k + 𝔯m. Then φ[k] = −q, W = Y ⊕ ∑_{i=2}^{r}(Ke_{i} + Kg_{i}), Λ = N + ∑_{i=2}^{r}(𝔯e_{i} + 𝔯g_{i}), and d_{0}(Y) = qd_{0}(φ). Since {K, qd_{0}(φ)} is a division algebra, (Y, φ) is anisotropic, and so has a unique maximal lattice M as noted in §3.7. Put Λ′ = M + ∑_{i=2}^{r}(𝔯e_{i} + 𝔯g_{i}). Then clearly Λ′ is the unique maximal lattice in W containing Λ.

To prove the remaining part, we assume that q𝔤 = 𝔭, 𝔡 = 𝔯, and d_{0}(φ) = N_{K/F} (K^{×}). Since N ⊂ M and [M̂: M] = N(𝔭)^{2} = [N̂ : N], we obtain M = N, and so Λ is maximal. We easily see that C(L)∩U^{φ}(W) ⊂ C(Λ). Let γ ∈ C(Λ).

Then Lγ is a maximal lattice in V containing Λ, so that ∑_{i=2}^{r}(𝔯e_{i} + 𝔯g_{i}) ⊂ Lγ. By [S2, Lemma 4.9 (i)] we have Lγ = J +∑_{i=2}^{r}(𝔯e_{i}+𝔯g_{i}) with a maximal lattice J in Km + Ke_{1} + Kg_{1}. Since m ∈ J, from (1.9) we see that φ(m, J) = 𝔯, and so J = 𝔯m + H with a maximal lattice H in Ke_{1} + Kg_{1} as shown in the proof of Lemma 3.4. Now h = hγ ∈ Lγ and k ∈ Λ = Λγ; thus h, k ∈ J. Put H_{0} = 𝔯e_{1} + 𝔯g_{1}. By Lemma 1.10 (ii), H = H_{0}α with α ∈ SU^{φ}(Ke_{1} + Kg_{1}). By (3.5) we can take α in the form (3.6). Identify GL(Ke_{1} + Kg_{1}) with GL_{2}(K) with respect to the basis {e_{1}, g_{1}}. Replacing α by an element of C(H_{0})α, we

can put α = [ \begin{matrix}a^{−1} & b \\ 0 & a \end{matrix} ] with a ∈ F^{×} and b ∈ F. Put z = a^{−1}e_{1} + bg_{1} and

w = ag_{1}; then H = 𝔯z + 𝔯w. Since δe_{1} = uk − u^{ρ}h ∈ H and qδg_{1} = h − k ∈ H, we see that 𝔭 ⊂ a𝔤 ⊂ 𝔤, and so we may assume that a = 1 or a = q. Now φ(z, e_{1}) ∈ 𝔯, and so b ∈ 𝔤. Thus if a = 1, then H = H_{0}. Suppose a = q; put

α_{b} = [ \begin{matrix}a^{−1} & b \\ 0 & a \end{matrix} ] . Then H_{0}α_{b} = H_{0}α_{b′} if and only if b − b′ ∈ 𝔭. Thus there

exist exactly N(𝔭) + 1 maximal lattices in Ke_{1} + Kg_{1} containing h and k, and so there exist at most N(𝔭) + 1 lattices of the form Lγ with γ ∈ C(Λ). This shows that

(3.11) [C(Λ) : C(L) ∩ U^{φ}(W)] ≤ N(𝔭) + 1.

To show that this is actually an equality, define the symbols ℓ, j, ω, and 𝔒 as in Lemma 1.3, (1.6), and §3.7, with Y as the space V there. In the proof of Lemma 1.3 take φ_{0} = diag[ζ, −q] and s = ζ^{−1}q. Then the identification of V with K_{2}^{1} in the proof of Lemma 1.3 (which is unrelated to the above identification of GL(Ke_{1} + Kg_{1}) with K_{2}^{2}) identifies m with (1, 0) and k with (0, 1), so that m here equals ℓ of Lemma 1.3; also, M = ℓj(𝔒) = mj(𝔒). Let β be an element of 𝔒 such that ββ^{ι} = 1; define ξ ∈ SU^{φ}(V) by ξ = j(β) on Y and xξ = x for every x ∈ Kh+∑_{i=2}^{r}(Ke_{i}+Kg_{i}). Then ξ ∈ C^{1}(Λ). Put β = c+dω with c, d ∈ 𝔯. Then ℓξ = cℓ + dk and kξ = sd^{ρ}ℓ + c^{ρ}k. Since δe_{1} = uk − u^{ρ}h and qδg_{1} = h − k, we have qδg_{1}ξ = (1 − c^{ρ})e_{1} − sd^{ρ}ℓ + q(u − c^{ρ}u^{ρ})g_{1}. Thus ξ ∈ C^{1}(L) if and only if c − 1 ∈ 𝔭𝔯, which is so if and only if β − 1 ∈ 𝔓. Since ββ^{ι} = 1, this shows that there exist at least N(𝔭) + 1 different Lγ with γ ∈ C^{1}(Λ), and so

N(𝔭) + 1 ≤ [C^{1}(Λ) : C^{1}(L) ∩ U^{φ}(W)].

This combined with (3.11) proves that

[C^{1}(Λ) : C^{1}(L) ∩ U^{φ}(W)] = [C(Λ) : C(L) ∩ U^{φ}(W)] = N(𝔭) + 1,

which completes the proof.

We insert here the classification of the structures (V, sφ) with s ∈ F^{×}. If K = F × F, the matter is settled in §1.8.

Proposition 3.15. (i) If F is a local field and dim(V) is even, then (V, φ) is isomorphic to (V, sφ) for every s ∈ F^{×}.

(ii) Suppose F is an arbitrary field and dim(V) is odd; let s ∈ F^{×}; then (V, φ) is isomorphic to (V, sφ) if and only if s ∈ N_{K/F} (K^{×}).

Proof. The first assertion is included in Lemma 1.6. The second assertion can be proved easily in the same manner as for [S3, Theorem 7.13] in the even-dimensional case.

4. Hermitian Diophantine equations over a global field

4.1. Throughout this section we assume that F is an algebraic number field and K is a quadratic extension of F; we fix a hermitian space (V, φ) and use the notation of §2.1. For an 𝔯-lattice L in V we put

(4.1)  Γ (L) = {α ∈ U^{φ}(V) | Lα = L},  Γ^{1}(L) = Γ (L) ∩ SU^{φ}(V).

Given h ∈ V such that φ[h] ≠ 0, put W = (Kh)^{⊥}. We view U^{φ}(W) as a subgroup of U^{φ}(V), and SU^{φ}(W) as a subgroup of SU^{φ}(V) as explained in §1.1. For σ ∈ U^{φ}(V)_{A}, the symbol hσ is meaningful as an element of V_{A}.

Theorem 4.2. If dim(V) > 1, then #{Λ[q]/Γ^{1}(Λ)} < ∞ for every q ∈ F^{×} and every 𝔯-lattice Λ in V.

Proof. Assuming dim(V) > 1 and Λ[q] ≠ ∅, take h ∈ Λ[q] and define W as above; put G = SU^{φ}(V), H = SU^{φ}(W), D = {x ∈ G_{A} | Λx = Λ}, and D_{v} = G_{v} ∩D for v ∈ h. Then Γ^{1}(Λ) = G∩D and D_{v} = C^{1}(Λ_{v}). By Theorem 3.3 and (1.5) we have Λ_{v}[q] = ⋃_{α∈X_{v}} hαD_{v} with a finite subset X_{v} of G_{v}. By Theorem 3.2 (iii) and (iv) we can take X_{v} = {1} if Λ_{v} is maximal, v is not ramified in K, and q ∈ 𝔤_{v}^{×}. Thus X_{v} = {1} for almost all v ∈ h. Put X = ∏_{v∈h} X_{v}. This is a finite subset of G_{h}. For each ξ ∈ X we can find a finite subset E_{ξ} of H_{h} such that H_{A} = ⋃_{ε∈E_{ξ}} Hε(H_{A} ∩ ξDξ^{−1}). Then H_{A}ξD = ⋃_{ε∈E_{ξ}} HεξD. For each (ε, ξ) such that G ∩ εξD ≠ ∅ pick β_{ε,ξ} ∈ G ∩ εξD. Now let k ∈ Λ[q]. Then k = hξζ for some ξ ∈ X and ζ ∈ D. On the other hand k = hα with α ∈ G by (1.5). Then αζ^{−1}ξ^{−1} ∈ H_{A}, so that α ∈ H_{A}ξD. Thus α ∈ HεξD for some ε ∈ E_{ξ}. Then α ∈ Hβ_{ε,ξ}D∩G = Hβ_{ε,ξ}Γ^{1}(Λ), and hence k = hα ∈ hb_{ε,ξ}Γ^{1}(Λ). Since the b_{ε,ξ} form a finite set, we obtain our theorem.

4.3. We now fix a maximal lattice L in V, and put

(4.2) C = {γ ∈ U^{φ}(V)_{A} | Lγ = L}, C^{1} = C ∩ SU^{φ}(V)_{A}.

We are going to state our main theorem with respect to a pair (G, H) belonging to the following two types of objects:

Type U: G = U^{φ}(V) and H = U^{φ}(W);

Type SU: G = SU^{φ}(V) and H = SU^{φ}(W).

Here W = (Kh)^{⊥} with a fixed h ∈ V. For a subset S of U^{φ}(V)_{A} the symbol hS is meaningful as a subset of V_{A}. Therefore V ∩ hS is a well-defined subset of V.

Theorem 4.4. Suppose dim(V) > 1. For a fixed h ∈ V such that φ[h] ≠ 0 put W = (Kh)^{⊥}, and take (G, H) of Type U or SU as above. Let D = D_{0}G_{a} with an open compact subgroup D_{0} of G_{h}. Then the following assertions hold.

(i) For y ∈ G_{A} we have H_{A} ∩ GyD ≠ ∅ if and only if V ∩ hDy^{−1} ≠ ∅.

(ii) Fixing y ∈ G_{A}, for every ε ∈ H_{A} ∩ GyD take α ∈ G so that ε ∈ αyD. Then the map ε ↦ hα gives a bijection of H\(H_{A} ∩ GyD)/(H_{A} ∩ D) onto (V ∩ hDy^{−1})/Δ_{y}, where Δ_{y} = G ∩ yDy^{−1}.

(iii) Take {y_{i}}_{i∈I} ⊂ G_{A} so that G_{A} = ⨆_{i∈I} Gy_{i}D, and put Γ_{i} = G ∩ y_{i}Dy_{i}^{−1}. Then

(4.3) #{H\H_{A}/(H_{A} ∩ D)} = ∑_{i∈I} #{(V ∩ hDy_{i}^{−1})/Γ_{i}}.

(iv) Let q = φ[h] and 𝔟 = φ(h, L). Then for every y ∈ U^{φ}(V)_{A}, we have

(4.4) V ∩ hCy^{−1} = (Ly^{−1})[q, 𝔟].

(v) Suppose moreover that dim(V) > 2 and the following condition is satisfied:

(4.5) If n is odd, then q_{v}𝔯_{v} ≠ 𝔟_{v}𝔟_{v}^{ρ} for every v ∈ h ramified in K.

Then for every y ∈ SU^{φ}(V)_{A} we have

(4.6) V ∩ hC^{1}y^{−1} = (Ly^{−1})[q, 𝔟].

Proof. Let y, ε, and α be as in (ii); then clearly hα ∈ V ∩ hDy^{−1}. If ηεζ ∈ βyD with η ∈ H, ζ ∈ H_{A} ∩ D, and β ∈ G, then β^{−1}ηα ∈ G ∩ yDy^{−1} = Δ_{y}, and hence hα = hηα ∈ hβΔ_{y}. Thus our map is well defined. Next let k ∈ V ∩ hDy^{−1}. Then k = hδy^{−1} with δ ∈ D, and moreover, by (1.5), k = hξ with ξ ∈ G. Then h = hξyδ^{−1}, so that ξyδ^{−1} ∈ H_{A}. Thus ξyδ^{−1} ∈ H_{A}∩GyD. This shows that k is the image of an element of H_{A} ∩ GyD. To prove that the map is injective, suppose ε ∈ αyD ∩ H_{A} and δ ∈ βyD ∩ H_{A} with α, β ∈ G, and hα = hβσ with σ ∈ Δ_{y}. Put ω = βσα^{−1}. Then hω = h, so that ω ∈ H. Since σ ∈ yDy^{−1}, we have βyD = βσyD = ωαyD, and hence δ ∈ βyD∩H_{A} = ωαyD ∩ H_{A} = ω(αyD ∩ H_{A}) = ω(εD ∩ H_{A}) = ωε(D ∩ H_{A}) ⊂ Hε(D ∩ H_{A}). This proves the injectivity, and completes the proof of (ii). At the same time we obtain (i).

Since H_{A} = ⨆_{i∈I}(H_{A} ∩ Gy_{i}D), we can derive (iii) immediately from (ii). As for (v), clearly V ∩hC^{1} ⊂ L[q, 𝔟]. Conversely, if x ∈ L[q, 𝔟], then x ∈ hC^{1}

by Theorem 3.2 (iii). Thus

(4.7) V ∩ hC^{1} = L[q, 𝔟].

If k ∈ V ∩ hC^{1}y^{−1}, then hC^{1}y^{−1} = kyC^{1}y^{−1}. Now φ(k, Ly^{−1}) = φ(h, L) = 𝔟. Taking k, yC^{1}y^{−1}, and Ly^{−1} in place of h, C^{1}, and L in (4.7), we obtain V ∩ kyC^{1}y^{−1} = (Ly^{−1})[q, 𝔟]. This proves (4.6) when V ∩hC^{1}y^{−1} ≠ ∅. To prove the remaining case, suppose ℓ ∈ (Ly^{−1})[q, 𝔟]; then φ(ℓ_{v}y_{v}, L_{v}) = 𝔟_{v} = φ(h, L)_{v} for every v ∈ h, and so, by Theorem 3.2 (iii), ℓy ∈ hC^{1}. Taking ℓ, yC^{1}y^{−1}, and Ly^{−1} in place of h, C^{1}, and L in (4.7), we obtain ∅ ≠ (Ly^{−1})[q, 𝔟] = V ∩ ℓyC^{1}y^{−1} = V ∩ hC^{1}y^{−1}. This shows that (Ly^{−1})[q, 𝔟] = ∅ if V ∩ hC^{1}y^{−1} = ∅, and hence (4.6) holds for every y ∈ G_{A}. This proves (v). Assertion (iv) can be proved in the same way.

In view of (i) we can restrict the indices i on the right-hand side of (4.3) to those for which H_{A} ∩ Gy_{i}D ≠ ∅. If I′ denotes the set of all such i’s, then H_{A} = ⨆_{j∈I′}(H_{A} ∩ Gy_{j}D).

Combining (4.3) and (4.4), we obtain, for (G, H) of type U, an equality

(4.8) #{H\H_{A}/(H_{A} ∩ C)} = ∑_{i∈I} #{(Ly_{i}^{−1})[q, 𝔟]/Γ_{i}},

where {y_{i}} is such that G_{A} = ⨆_{i∈I} Gy_{i}C and Γ_{i} = G ∩ y_{i}Cy_{i}^{−1}. We can state a similar formula for (G, H) of type SU when n > 2 and (4.5) is satisfied. Formula (4.8) connects the class number of H with respect to H_{A} ∩ C to the solutions h of the equation φ[h] = q under the condition φ(h, Ly_{i}^{−1}) = 𝔟.

5. Nonscalar hermitian Diophantine equations

5.1. So far we discussed the equation φ[h] = q with a scalar q. We can formulate a similar problem with nonscalar q, which can be stated in terms of matrices as follows. We take F to be local or global. Given q^{∗} = q ∈ GL_{m}(K) and φ^{∗} = φ ∈ GL_{n}(K), we consider the solutions h ∈ K_{n}^{m} of the equation hφh^{∗} = q. Here and throughout this section we assume n > m > 0. More intrinsically, take (V, φ) as before and take also (X, q) with a nondegenerate hermitian form q on a free K-module X of dimension m. We consider h ∈ Hom(X, V) such that φ[xh] = q[x] for every x ∈ X. Since q is nondegenerate, h must be injective. To simplify our notation, for every k ∈ Hom(X, V) we denote by φ[k] the hermitian form on X defined by φ[k][x] = φ[xk] for every x ∈ X. Then our problem concerns the solutions h ∈ Hom(X, V) of the equation φ[h] = q for a fixed q. If m = 1 and X = K, then q ∈ F^{×}, and an element h of V defines an element of Hom(K, V) that sends c to ch for c ∈ K, and every element of Hom(K, V) is of this type. Thus the problem about φ[h] = q with q ∈ F^{×} is the one-dimensional special case. Let h be an element of Hom(X, V) such that rank(φ[h]) = m. Then

(5.1) {x ∈ Hom(X, V) | φ[x] = φ[h]} = h · SU^{φ}.

This is similar to (1.5), and follows easily from the Witt theorem in the unitary case. Though we take V to be coordinate-free, it is practical to take X to be K_{m}^{1}, and so take q to be a hermitian element of GL_{m}(K), and q[x] = xqx^{∗} for x ∈ K_{m}^{1}. For h ∈ Hom(X, V) and 1 ≤ i ≤ m we define “the i-th row” of h to be the element h_{i} of V determined by ah = ∑_{i=1}^{m} a_{i}h_{i} for a = (a_{i})_{i=1}^{m} ∈ K_{m}^{1} = X.

We first prove a local finiteness result that generalizes Theorem 3.3.

Theorem 5.2. Suppose F is a local field; let Λ be an 𝔯-lattice in Hom(X, V) and let D = {γ ∈ SU^{φ}(V) | Λγ = Λ}. Then, given h ∈ Hom(X, V) ∩ Λ such that φ[h] is nondegenerate, there exists a finite subset A of SU^{φ}(V) such that

(5.2) { x ∈ Λ | φ[x] = φ[h] } = ⨆_{α∈A} hαD.

Moreover, suppose K = F × F or K is a field unramified over F; suppose also that φ[h] ∈ GL_{m}(𝔯) and Λ = {λ ∈ Hom(X, V) | 𝔯_{m}^{1}λ ⊂ L} with a maximal lattice L in V. Then we can take A = {1}.

Proof. The first part is Theorem 3.3 if m = 1, and so we assume m > 1 and prove (5.2) by induction on m. Put q = φ[h]. Changing (h, q, Λ) for (ch, cqc^{∗}, cΛ) with a suitable c ∈ GL_{m}(K), we may assume that q = diag[a, η] with a ∈ F^{×} and η^{∗} = η ∈ GL_{m−1}(K). Also we may assume that Λ = {κ ∈ Hom(X, V) | Mκ ⊂ L} with M = 𝔯_{m}^{1} and an 𝔯-lattice L in V. Then D = {α ∈ SU^{φ}(V) | Lα = L}. If x ∈ Λ and φ[x] = q, then x_{1}, h_{1} ∈ L and φ[x_{1}] = φ[h_{1}] = a, and hence by Theorem 3.3 there exists a finite subset B of L such that such an x_{1} belongs to ⋃_{b∈B} bD and φ[b] = a for every b ∈ B. Suppose x_{1} = bγ with b ∈ B and γ ∈ D. Put W_{b} = (Kb)^{⊥}, y = xγ^{−1} and z =[y_{i}]_{i=2}^{m}. Then y_{1} =b and φ[y]=φ[h], so that φ[z]=η, and y_{i} ∈W_{b} for i > 1. We can view z as an element of Hom(K_{m−1}^{1}, W_{b}). Then 𝔯_{m−1}^{1}z ⊂ L ∩ W_{b}. Put E ={ε∈SU^{φ}(W_{b}) | (L ∩ W_{b})ε=L ∩ W_{b}}. By induction there exists a finite subset U_{b} of Hom(K_{m−1}^{1}, W_{b}) such that

{z ∈ Hom(K_{m−1}^{1}, W_{b}) | 𝔯_{m−1}^{1}z ⊂ L ∩ W_{b}, φ[z]=η } = ⨆_{u∈U_{b}} uE.

We can find a finite subset S of E such that E = ⨆_{σ∈S} σ(D ∩ E). Then y =

[ \begin{matrix}b \\ z \end{matrix} ] = [ \begin{matrix}b \\ uστ \end{matrix} ] with u ∈ U_{b}, σ ∈ S, and τ ∈ D ∩ E. Thus x = [ \begin{matrix}b \\ uσ \end{matrix} ] τγ, and

τγ ∈ D. This shows that x ∈ ⨆_{k∈P} kD with a finite subset P of the left-hand side of (5.2), as the elements (b, u, σ) form a finite set. By (5.1), for each k ∈P there exists an element α of SU^{φ}(V) such that k = hα. This proves the first assertion.

Next suppose that the conditions on K, q, Λ, and φ as in the second assertion are satisfied. Take an 𝔯-basis of L and identify V, L, and φ with K_{n}^{1}, 𝔯_{n}^{1}, and a hermitian matrix with respect to that basis. Given ℓ ∈ L = 𝔯_{n}^{1}, put z = q^{−1}hφℓ^{∗} and y = ℓ − z^{∗}h. Since φ ≺ 𝔯 and q ∈ GL_{m}(𝔯), we see that y ∈ L, and for every w ∈ X we have whφy^{∗} = 0, so that y ∈ (Xh)^{⊥}. Put M = 𝔯_{m}^{1} and Y = (Xh)^{⊥}. Then V = Xh ⊕ Y and L = Mh ⊕ (L ∩ Y). Suppose φ[k]=q with k ∈Λ. Then similarly L=Mk ⊕ (L ∩ Z) with Z =(Xk)^{⊥}. Since L is maximal, Mh resp. Mk is maximal in Xh resp. Xk, and L ∩ Y resp. L ∩ Z is maximal in Y resp. Z. By (5.1) there exists an element γ ∈ SU^{φ}(V) such that hγ = k. Then Mhγ = Mk, Y γ = Z, and (L ∩ Y)γ is 𝔯-maximal in Z, so that by Lemma 3.4 (i), (L ∩ Y)γε = L ∩ Z with some ε ∈ U^{φ}(Z). Define α ∈ GL(V) by α = γ on Xh and α = γε on Y. Then α ∈ U^{φ}(V), Lα = L, and hα = k. Since det(α) ∈ 𝔯^{×}, we can find an element ξ of U^{φ}(Z) such that (L ∩ Z)ξ = L ∩ Z and det(ξ) = det(α)^{−1}. This is clear if K = F × F; see §3.6. If K is a field unramified over F, then the fact is included in Lemma 1.9. Extend ξ to an element of U^{φ}(V) by putting xkξ = xk for x ∈ X. Then αξ ∈ SU^{φ}(V), hαξ = k, and Lαξ = L. Clearly Λαξ = Λ, and hence we obtain (5.2) with A={1}. This completes the proof.

Next we prove a generalization of Theorem 4.2, which is a global version of the above theorem.

Theorem 5.3. Suppose that F is an algebraic number field; let Λ be an 𝔯lattice in Hom(X, V), Γ = { γ ∈ SU^{φ}(V) | Λγ = Λ }, and T_{q} = { x ∈ Λ | φ[x] = q } with q^{∗} = q ∈ GL_{m}(K). Then T_{q}/Γ is a finite set.

Proof. We assume the existence of h ∈ T_{q}. Put W = (Xh)^{⊥}, G = SU^{φ}(V), H = SU^{φ}(W), M = 𝔯_{m}^{1}, D = {γ ∈ G_{A} | Λγ = Λ}, and D_{v} = D ∩ G_{v} for v ∈ h. We identify H with {α ∈ G | hα = h}. Fix a maximal lattice L in V. By Theorem 5.2, for each v ∈ h there exists a finite subset E_{v} of G_{v} such that

{x ∈ Λ_{v} | φ[x] = q } = ⨆_{ε∈E_{v}} hεD_{v}.

Now for almost all v ∈ h we have Λ_{v} = {γ ∈ Hom(X_{v}, V_{v}) | M_{v}γ ⊂ L_{v}}, L_{v} is maximal, v is unramified in K, and q ∈ GL_{m}(𝔯_{v}). Therefore, by Theorem 5.2, we can take E_{v} = {1} for almost all v ∈ h. Consequently we can find a finite subset E of G_{h} such that T_{q} ⊂ ⋃_{η∈E} hηD. If x ∈ T_{q}, then x ∈ hG by (5.1). Thus x = hα = hηδ with α ∈ G, η ∈ E, and δ ∈ D. We have αδ^{−1}η^{−1} ∈ H_{A}, and hence α ∈ H_{A}ηD. For each η ∈ E we can find a finite subset Z_{η} of H_{h} such that H_{A} = ⨆_{ζ∈Z_{η}} Hζ(H_{A} ∩ ηDη^{−1}). Then H_{A}ηD = ⋃_{ζ∈Z_{η}} HζηD, and hence α ∈ ⋃_{η,ζ}(G ∩ HζηD) = ⋃_{η,ζ} H(G ∩ ζηD). For each (ζ, η) such that G ∩ ζηD ≠ ∅, pick any β ∈ G ∩ ζηD. Then G ∩ ζηD = G ∩ βD = βΓ. Let B be the set of such β’s chosen for each (ζ, η). Then α ∈ ⋃_{β∈B} HβΓ, and thus hα ∈ ⋃_{β∈B} hβΓ, which proves our theorem.

Theorem 5.4. Suppose that F is an algebraic number field. With a fixed h ∈ Hom(X, V) such that rank(φ[h]) = m, put q = φ[h], W = (Xh)^{⊥}, G = U^{φ}(V), H = U^{φ}(W), and 𝒱 = Hom(X, V). Let D be an open subgroup of G_{A} containing G_{a} such that D ∩ G_{h} is compact, and let G_{A} = ⨆_{i∈I} Gy_{i}D. Then assertions (i), (ii), and (iii) of Theorem 4.4 are valid if we take the symbols h, G, H, and D there to be those of the present setting, and replace V there by 𝒱 . In particular we have

(5.3) #{H\H_{A}/(H_{A} ∩ D)} = ∑_{i∈I} #{(𝒱 ∩ hDy_{i}^{−1})/Γ_{i}},

where Γ_{i} = G∩y_{i}Dy_{i}^{−1}. The same is true with G = SU^{φ}(V) and H = SU^{φ}(W).

Proof. We can repeat the proof of Theorem 4.4 with obvious modifications.

Theorem 5.5. In the setting of Theorem 5.4 with G = SU^{φ}(V) and H = SU^{φ}(W) suppose that n − m > 1 and neither G_{a} nor H_{a} is compact. Let k be an element of Hom(X, V) such that k = hγ_{v} for every v ∈ h with some (γ_{v})_{v∈h} ∈ D ∩ G_{h}. Then there exists an element α ∈ G ∩ D such that k = hα. In particular, if m = 1 and (4.5) is satisfied, then #{L[q, 𝔟]/Γ^{1}(L)} ≤ 1 for every maximal lattice L in V.

Proof. By our assumptions, strong approximation holds on G and H, and so we have G_{A} = GD and H_{A} = H(H_{A} ∩ D). Thus we can take {y_{i}}_{i∈I} = {1}. Therefore (5.3) implies that #{(𝒱 ∩ hD)/(G ∩ D)} = 1, which gives the first assertion. This combined with (4.6) proves the second assertion.

5.6. Before proceeding further, let us recall the notion of the mass of an algebraic group G with respect to an open subgroup D of G_{A} containing G_{a} and such that G_{h} ∩ D is compact. For simplicity here we take G to be U^{φ} or SU^{φ} and assume that U_{a}^{φ} is compact. For x ∈ G_{A} put Δ_{x} = G ∩ xDx^{−1} and ν(Δ_{x}) = [Δ_{x} : 1]^{−1}. Then the the mass of G with respect to D is defined by

(5.4) 𝔪(G, D) = ∑_{b∈ℬ} ν(Δ_{b}), ℬ = G\G_{A}/D.

For this the reader is referred to [S2, (10.9.4), (24.1.1), (24.1.2)]. If D′ is a subgroup of G_{A} of the same type as D, then from [S2, Lemma 24.2] we obtain

(5.5) [D : D ∩ D′]𝔪(G, D) = 𝔪(G, D ∩ D′) = [D′ : D ∩ D′]𝔪(G, D′).

Theorem 5.7. In the setting of Theorem 5.4, suppose that G_{a} is compact. Then for every y ∈ G_{A} we have

(5.6) ν(Δ_{y})#{𝒱 ∩ hDy^{−1}} = ∑_{ε∈ℰ}ν(Δ_{ε}),

where ℰ = H\(H_{A} ∩ GyD)/(H_{A} ∩ D) and Δ_{x} = H ∩ xDx^{−1}. Moreover, let G_{A} = ⨆_{i∈I} Gy_{i}D and Γ_{i} = G ∩ y_{i}Dy_{i}^{−1}; then

(5.7)  ∑_{i∈I} ν(Γ_{i})#{𝒱 ∩ hDy_{i}^{−1}} = 𝔪(H, H_{A} ∩ D).

Proof. To prove (5.6), we may assume that H_{A} ∩ GyD ≠ ∅. For ε ∈ ℰ take α_{ε} ∈ G so that ε ∈ α_{ε}yD. Then H ∩ α_{ε}Δ_{y}α_{ε}^{−1} = H ∩ α_{ε}yDy^{−1}α_{ε}^{−1} = H ∩ εDε^{−1} = Δ_{ε}. Now 𝒱 ∩ hDy^{−1} = ⨆_{ε∈ℰ} hα_{ε}Δ_{y} by the part of Theorem 5.4 corresponding to Theorem 4.4 (ii). For γ, γ′ ∈ Γ (Λ) we have hα_{ε}γ = hα_{ε}γ′ if and only if α_{ε}γ′γ^{−1}α_{ε}^{−1} ∈ H, that is, γ′γ^{−1} ∈ α_{ε}^{−1}Hα_{ε} ∩ Δ_{y} = α_{ε}^{−1}Δ_{ε}α_{ε}, so that

#{hα_{ε}Δ_{y}} = [Δ_{y} : α_{ε}^{−1}Δ_{ε}α_{ε}] = ν(Δ_{ε})/ν(Δ_{y}).

Therefore we obtain (5.6). Next, let ℰ_{i} = H\(H_{A} ∩ Gy_{i}D)/(H_{A} ∩ D). Then H_{A} = ⨆_{i∈I}(H_{A} ∩ Gy_{i}D) = ⨆_{i∈I} ⨆_{ε∈ℰ_{i}} Hε(H_{A} ∩ D), and so 𝔪(H, H_{A} ∩ D) = ∑_{i∈I} ∑_{ε∈ℰ_{i}} ν(Δ_{ε}), which combined with (5.6) proves (5.7).

Corollary 5.8. Define C and C^{1} by (4.2) with a maximal lattice L in V; take (G, H) of type U as in Theorem 4.4; suppose that G_{a} is compact. Let G_{A} = ⨆_{i∈I} Gy_{i}C, L_{i} = Ly_{i}^{−1}, and Γ_{i} = Γ (L_{i}). Then

(5.8)  ∑_{i∈I} ν(Γ_{i})#{L_{i}[q, 𝔟]} = 𝔪(H, H_{A} ∩ C),

where q = φ[h] and 𝔟 = φ(h, L). This is valid for (G, H) of type SU if we replace C and Γ (··) by C^{1} and Γ^{1}(··), provided n > 2 and (4.5) is satisfied.

Proof. Take m = 1 and D = C in Theorem 5.7. Combining (5.7) with (4.4), we obtain (5.8). The case of SU^{φ} follows similarly from (4.6).

5.9. Formulas (5.7) and (5.8) are similar to, but different from, the formula of Siegel about ∑_{i} ν(Γ_{i})#{L_{i}[q]}. We already explained in [S3, §13.13] the main differences between our formulas in the orthogonal case given in that book and that of Siegel. In principle, our comments there apply to the present unitary case.

Now in [S2, Theorem 24.4] we gave an exact formula for 𝔪(G, D) for G = U^{φ} and a certain type of D, under the condition that if n is odd, then d_{0}(φ) is represented by an element of 𝔤^{×}. The group H_{A}∩C in (5.8) does not necessarily belong to the types of D there, but we can compute [D : H_{A} ∩ C] by means of Lemma 3.14 under some conditions on (q, 𝔟). Then we obtain 𝔪(H, H_{A} ∩ C) from [S2, Theorem 24.4] by (5.5).

Proposition 5.10. In the setting of Theorem 5.4 suppose that n − m is odd. Then the structure (W, det(q)φ) depends only on φ and the indices of q at the real archimedean primes of F ramified in K.

Proof. Let ψ be the restriction of φ to W. Then we can easily verify that d_{0}( det(q)ψ) = det(q)^{n−m}d_{0}(ψ) = (−1)^{n−1}d_{0}(φ) as m − n is odd. This combined with Theorem 2.2 (i) proves our proposition.

This is an analogue of the fact concerning a quadratic form in even dimension with square discriminant given in [S4, Theorem 1.12].

We insert here some results about the relationship between various invariants associated with U^{φ} and those with SU^{φ}.

Proposition 5.11. Let D be an open subgroup of U_{A}^{φ} containing U_{a}^{φ} and such that U_{h}^{φ} ∩D is compact; put P = {x ∈ K^{×} | xx^{ρ} = 1} and D^{1} = D∩SU_{A}^{φ}. Then U^{φ}SU_{A}^{φ}D is a normal subgroup of U_{A}^{φ} and (5.9a) [U_{A}^{φ} : U^{φ}SU_{A}^{φ}D] = [P_{A} : P det(D)],

(5.9b) #(U^{φ}\U_{A}^{φ}/D) ≤ ∑_{x∈Ξ} #(SU^{φ}\SU_{A}^{φ}/xD^{1}x^{−1}),

where Ξ = U_{A}^{φ}/U^{φ}SU_{A}^{φ}D. Moreover, if U_{a}^{φ} is compact, then

(5.9c) 𝔪(U^{φ}, D) ≤ [P_{A} : P det(D)] · 𝔪(SU^{φ}, D^{1}).

Furthermore, if P ∩ det(D) = det(U^{φ} ∩ yDy^{−1}) for every y ∈ U_{A}^{φ}, then the equality holds in (5.9b), and

(5.9d) #(P ∩ det(D))𝔪(U^{φ}, D) = [P_{A} : P det(D)]𝔪(SU^{φ}, D^{1}).

Proof. Since P = det(U^{φ}), we can easily show that

(5.10) U^{φ}SU_{A}^{φ}Dx = {y ∈ U_{A}^{φ} | det(y) ∈ P det(Dx)}

for every x ∈ U_{A}^{φ}. This shows that U^{φ}SU_{A}^{φ}D is a normal subgroup of U_{A}^{φ} and U_{A}^{φ}/U^{φ}SU_{A}^{φ}D is isomorphic to P_{A}/[P det(D)], as P_{A} = det(U_{A}^{φ}). Thus we obtain (5.9a); we also see that U^{φ}SU_{A}^{φ}\U_{A}^{φ}/D can be identified with U_{A}^{φ}/U^{φ}SU_{A}^{φ}D. Given x ∈ U_{A}^{φ}, take B_{x} ⊂ SU_{A}^{φ} so that SU_{A}^{φ} = ⨆_{b∈B_{x}} SU^{φ}bxD^{1}x^{−1}. Then we have U^{φ}SU_{A}^{φ}xD = ⋃_{b∈B_{x}} U^{φ}bxD, and hence U_{A}^{φ} = ⋃_{x∈Ξ} ⋃_{b∈B_{x}} U^{φ}bxD. From this we obtain (5.9b). To prove (5.9c), put Γ_{x} = U^{φ} ∩ xDx^{−1} and Γ_{x}^{1} = SU^{φ} ∩ xD^{1}x^{−1} for x ∈ U_{A}^{φ}. Then 𝔪(U^{φ}, D) ≤ ∑_{x∈Ξ}∑_{b∈B_{x}} ν(Γ_{bx}) ≤ ∑_{x∈Ξ}∑_{b∈B_{x}} ν(Γ_{bx}^{1}) = ∑_{x∈Ξ} 𝔪(SU^{φ}, xD^{1}x^{−1}). Now formula (5.5) shows that 𝔪(SU^{φ}, D^{1}) depends only on the measure of D^{1}. (If U_{a}^{φ} is not compact, we have to consider the measure of D_{h}^{1}.) Since 𝔪(SU^{φ}, xD^{1}x^{−1}) = 𝔪(SU^{φ}, D^{1}), we obtain (5.9c).

Suppose P ∩ det(D) = det(Γ_{y}) for every y ∈ U_{A}^{φ}. Suppose also that b′x = abxd for a ∈ U^{φ}, d ∈ D, and b, b′ ∈ B_{x}. Then det(a) = det(d^{−1}) ∈ P ∩ det(D) = det(Γ_{bx}), and so det(a) = det(c) with c ∈ Γ_{bx}. Put e = x^{−1}b^{−1}cbx. Then e ∈ D, det(ed) = 1, and b′x = abxd = ac^{−1}bxed ∈ SU^{φ}bxD^{1}. Thus b′ = b. This shows that U^{φ}SU_{A}^{φ}xD = ⨆_{b∈B_{x}} SU^{φ}bxD, from which we obtain the equality in (5.9b). Also, ν(Γ_{bx}^{1})/ν(Γ_{bx}) = [Γ_{bx} : Γ_{bx}^{1}] = #( det(Γ_{bx})) = #(P ∩ det(D)), and so

#(P ∩ det(D))𝔪(U^{φ}, D) = ∑_{x∈Ξ} ∑_{b∈B_{x}} ν(Γ_{bx}^{1})

= ∑_{x∈Ξ} 𝔪(SU^{φ}, xD^{1}x^{−1}) = #(Ξ) · 𝔪(SU^{φ}, D^{1}),

which is (5.9d).

Department of Mathematics, Princeton University, Princeton, NJ

References

[E] M. Eichler, Zur Zahlentheorie der Quaternionen-Algebren, J. reine angew. Math. 195 (1956), 127–151.

[L] W. Landherr, Äquivalenz Hermitscher Formen über einem beliebigen algebraischen Zahlkörper, Abh. Math. Sem. Hamb., 11 (1936), 245-248.

[S1] G. Shimura, Arithmetic of unitary groups, Ann. of Math., 79 (1964), 369–409 (=Collected Papers, I, 473-513).

[S2] G. Shimura, Euler Products and Eisenstein series, CBMS Regional Conference Series in Mathematics, No. 93, American Mathematical Society, 1997.

[S3] G. Shimura, Arithmetic and analytic theories of quadratic forms and Clifford groups, Mathematical Surveys and Monographs, vol. 109, American Mathematical Society, 2004.

[S4]G. Shimura, Quadratic Diophantine equations and orders in quaternion algebras, Amer. J. Math. 128 (2006), 481–518.

[S5] G. Shimura, Classification, construction, and similitudes of quadratic forms, Amer. J. Math. 128 (2006), 1521–1552.

[S6] G. Shimura, Integer-valued quadratic forms and quadratic Diophantine equations, Documenta Mathematica 11, (2006), 333-367.

[S7] G. Shimura, Quadratic Diophantine equations, the class number, and the mass formula, Bull. Amer. Math. Soc. 43 (2006), 285-304.

Goro Shimura

Department of Mathematics Princeton University Princeton NJ 08544–1000 USA goro@Math.Princeton.EDU